View source for List of set identities and relations

{{short description|Equalities for combinations of sets}}
This article lists [[Mathematics|mathematical]] properties and laws of [[Set (mathematics)|sets]], involving the set-theoretic [[Operation (mathematics)|operations]] of [[Union (set theory)|union]], [[Intersection (set theory)|intersection]], and [[Complement (set theory)|complementation]] and the [[Binary relation|relations]] of set [[Equality (mathematics)|equality]] and set [[Subset|inclusion]]. It also provides systematic procedures for evaluating expressions, and performing calculations, involving these operations and relations.

The [[Binary operation|binary operation]]s of set union (<math>\cup</math>) and intersection (<math>\cap</math>) satisfy many identities. Several of these identities or "laws" have well established names. 

==Notation==

Throughout this article, capital letters (such as <math>A, B, C, L, M, R, S,</math> and <math>X</math>) will denote sets. On the left hand side of an identity, typically,
* <math>L</math> will be the '''{{em|L}}{{hsp}}'''eft most set, 
* <math>M</math> will be the '''{{em|M}}'''{{hsp}}iddle set, and 
* <math>R</math> will be the '''{{em|R}}'''{{hsp}}ight most set.
This is to facilitate applying identities to expressions that are complicated or use the same symbols as the identity.<ref group=note>
For example, the expression <math>(M \setminus R) \setminus A</math> uses two of the same symbols (<math>M</math> and <math>R</math>) that appear in the identity 
<math display=block>(L \,\setminus\, M) \,\setminus\, R ~=~ (L \,\setminus\, R) \,\setminus\, (M \,\setminus\, R)</math>
but they refer to different sets in each expression. 
To apply this identity to <math>(M \setminus R) \setminus A,</math> substitute <math>\text{Left set} := M,\;</math> <math>\text{Middle set} := R,\;</math> and <math>\text{Right set} := A</math> (since these are the left, middle, and right sets in <math>(M \setminus R) \setminus A</math>) to obtain: 
<math display=block>\begin{alignat}{4}
(M \setminus R) \setminus A 
&= (\text{Left set } &&\setminus \text{Right set}&&) &&\setminus (\text{Middle set } &&\setminus \text{Right set}) \\
&= (M &&\setminus A&&) &&\setminus (R &&\setminus A). \\
\end{alignat}</math>
For a second example, this time applying the identity to <math>((M \cap R \setminus L) \setminus (A \triangle L)) \setminus L,</math> is now given. The identity <math display=inline>(L \setminus M) \setminus R = (L \setminus R) \setminus (M \setminus R)</math> can be applied to <math>((M \cap R \setminus L) \setminus (A \triangle L)) \setminus L</math> by reading <math>L, M,</math> and <math>R</math> as <math>\text{Left}, \text{Middle},</math> and <math>\text{Right}</math> and then substituting <math>\text{Left} = (M \cap R \setminus L),</math> <math>\text{Middle} = (A \triangle L),</math> and <math>\text{Right} = L</math> to obtain:
<math display=block>\begin{alignat}{4}
((M \cap R \setminus L) \setminus (A \triangle L)) \setminus L 
&= (\text{Left } &&\setminus \text{Right}&&) &&\setminus (\text{Middle } &&\setminus \text{Right}) \\
&= ((M \cap R \setminus L) &&\setminus L&&) &&\setminus ((A \triangle L) &&\setminus L). \\
\end{alignat}</math>
</ref> 
For example, the identity 
<math display=block>(L \,\setminus\, M) \,\setminus\, R ~=~ (L \,\setminus\, R) \,\setminus\, (M \,\setminus\, R)</math>
may be read as:
<math display=block>(\text{Left set} \,\setminus\, \text{Middle set}) \,\setminus\, \text{Right set} ~=~ (\text{Left set} \,\setminus\, \text{Right set}) \,\setminus\, (\text{Middle set} \,\setminus\, \text{Right set}).</math>

===Elementary set operations===

For sets <math>L</math> and <math>R,</math> define:
<math display=block>\begin{alignat}{4}
L \cup R      &&~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{~ x ~:~ x \in L \;&&\text{ or }\;\, &&\; x \in R ~\} \\
L \cap R      &&~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{~ x ~:~ x \in L \;&&\text{ and }    &&\; x \in R ~\} \\
L \setminus R &&~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{~ x ~:~ x \in L \;&&\text{ and }    &&\; x \notin R ~\} \\
\end{alignat}</math>
and
<math display=block>L \triangle R ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{~ x ~:~ x \text{ belongs to exactly one of } L \text{ and } R ~\}</math>
where the {{em|[[Symmetric difference|symmetric difference]]}} <math>L \triangle R</math> is sometimes denoted by <math>L \ominus R</math> and equals:<ref name="Taylor 2019">{{Cite web|last=Taylor|first=Courtney|date=March 31, 2019|title=What Is Symmetric Difference in Math?|url=https://www.thoughtco.com/what-is-the-symmetric-difference-3126594|access-date=2020-09-05|website=ThoughtCo|language=en}}</ref><ref name="Weisstein 2020">{{Cite web|last=Weisstein|first=Eric W.|title=Symmetric Difference|url=https://mathworld.wolfram.com/SymmetricDifference.html|access-date=2020-09-05|website=mathworld.wolfram.com|language=en}}</ref>
<math display=block>\begin{alignat}{4}
L \;\triangle\; R 
~&=~ (L ~\setminus~ &&R) ~\cup~      &&(R ~\setminus~ &&L) \\
~&=~ (L ~\cup~      &&R) ~\setminus~ &&(L ~\cap~      &&R). 
\end{alignat}</math>

One set <math>L</math> is said to {{em|intersect}} another set <math>R</math> if <math>L \cap R \neq \varnothing.</math> Sets that do not intersect are said to be {{em|[[Disjoint sets|disjoint]]}}. 

The [[Power set|power set]] of <math>X</math> is the set of all subsets of <math>X</math> and will be denoted by
<math display=block>\wp(X) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{~ L ~:~ L \subseteq X ~\}.</math>

'''Universe set and complement notation'''

The notation 
<math display=block>L^\complement ~\stackrel{\scriptscriptstyle\text{def}}{=}~ X \setminus L.</math>
may be used if <math>L</math> is a subset of some set <math>X</math> that is understood (say from context, or because it is clearly stated what the superset <math>X</math> is). 
It is emphasized that the definition of <math>L^\complement</math> depends on context. For instance, had <math>L</math> been declared as a subset of <math>Y,</math> with the sets <math>Y</math> and <math>X</math> not necessarily related to each other in any way, then <math>L^\complement</math> would likely mean <math>Y \setminus L</math> instead of <math>X \setminus L.</math>

If it is needed then unless indicated otherwise, it should be assumed that <math>X</math> denotes the [[Universe (mathematics)|universe set]], which means that all sets that are used in the formula are subsets of <math>X.</math> 
In particular, the [[Complement (set theory)|complement of a set]] <math>L</math> will be denoted by <math>L^\complement</math> where unless indicated otherwise, it should be assumed that <math>L^\complement</math> denotes the complement of <math>L</math> in (the universe) <math>X.</math>

==One subset involved==

Assume <math>L \subseteq X.</math>

'''[[Identity element|Identity]]''':{{sfn|Monk|1969|pp=24-54}}

'''Definition''': <math>e</math> is called a '''left identity element''' of a binary operator <math>\,\ast\,</math> if <math>e \,\ast\, R = R</math> for all <math>R</math> and it is called a '''right identity element''' of <math>\,\ast\,</math> if <math>L \,\ast\, e = L</math> for all <math>L.</math> A left identity element that is also a right identity element if called an '''[[Identity element|identity element]]'''.

The empty set <math>\varnothing</math> is an identity element of binary union <math>\cup</math> and symmetric difference <math>\triangle,</math> and it is also a right identity element of set subtraction <math>\, \setminus:</math> 

<math display=block>\begin{alignat}{10}
L \cap X                  &\;=\;&& L &\;=\;&  X \cap L                  ~~~~\text{ where } L \subseteq X \\[1.4ex]
L \cup \varnothing        &\;=\;&& L &\;=\;& \varnothing \cup L        \\[1.4ex]
L \,\triangle \varnothing &\;=\;&& L &\;=\;& \varnothing \,\triangle L \\[1.4ex]
L \setminus \varnothing   &\;=\;&& L \\[1.4ex]
\end{alignat}</math>
but <math>\varnothing</math> is not a left identity element of <math>\, \setminus \,</math> since
<math display=block>\varnothing \setminus L = \varnothing</math> 
so <math display=inline>\varnothing \setminus L = L</math> if and only if <math>L = \varnothing.</math>

'''Idempotence'''{{sfn|Monk|1969|pp=24-54}} <math>L \ast L = L</math> and '''[[Nilpotent|Nilpotence]]''' <math>L \ast L = \varnothing</math>:

<math display=block>\begin{alignat}{10}
L \cup L                  &\;=\;&& L            && \quad \text{ (Idempotence)} \\[1.4ex]
L \cap L                  &\;=\;&& L            && \quad \text{ (Idempotence)} \\[1.4ex]
L \,\triangle\, L         &\;=\;&& \varnothing  && \quad \text{ (Nilpotence of index 2)} \\[1.4ex]
L \setminus L             &\;=\;&& \varnothing  && \quad \text{ (Nilpotence of index 2)} \\[1.4ex]
\end{alignat}</math>

'''Domination'''{{sfn|Monk|1969|pp=24-54}}/[[Zero element]]:

<math display=block>\begin{alignat}{10}
X \cup L                  &\;=\;&& X           &\;=\;& L \cup X ~~~~\text{ where } L \subseteq X \\[1.4ex]
\varnothing \cap L        &\;=\;&& \varnothing &\;=\;& L \cap \varnothing  \\[1.4ex]
\varnothing \times L      &\;=\;&& \varnothing &\;=\;& L \times \varnothing \\[1.4ex]
\varnothing \setminus L   &\;=\;&& \varnothing &\;\;& \\[1.4ex]
\end{alignat}</math>
but 
<math display=block>L \setminus \varnothing = L</math> 
so <math display=inline>L \setminus \varnothing = \varnothing \text{ if and only if } L = \varnothing.</math>

'''Double complement''' or '''[[Involution (mathematics)|involution]] law''':

<math display=block>\begin{alignat}{10}
X \setminus (X \setminus L)
&= L
&&\qquad\text{ Also written }\quad
&&\left(L^\complement\right)^\complement = L
&&\quad&&\text{ where } L \subseteq X \quad
\text{ (Double complement/Involution law)} \\[1.4ex]
\end{alignat}</math>

<math display=block>L \setminus \varnothing = L</math>
<math display=block>\begin{alignat}{4}
\varnothing 
&= L           &&\setminus L \\
&= \varnothing &&\setminus L \\
&= L           &&\setminus X ~~~~\text{ where } L \subseteq X \\
\end{alignat}</math>{{sfn|Monk|1969|pp=24-54}}

<math display=block>L^\complement = X \setminus L \quad \text{ (definition of notation)}</math>

<math display=block>\begin{alignat}{10}
L \,\cup (X \setminus L)
&= X
&&\qquad\text{ Also written }\quad
&&L \cup L^\complement = X
&&\quad&&\text{ where } L \subseteq X
\\[1.4ex]

L \,\triangle (X \setminus L)
&= X
&&\qquad\text{ Also written }\quad
&&L \,\triangle L^\complement = X
&&\quad&&\text{ where } L \subseteq X
\\[1.4ex]

L \,\cap (X \setminus L)
&= \varnothing
&&\qquad\text{ Also written }\quad
&&L \cap L^\complement = \varnothing
&&\quad&&
\\[1.4ex]
\end{alignat}</math>{{sfn|Monk|1969|pp=24-54}}

<math display=block>\begin{alignat}{10}
X \setminus \varnothing
&= X
&&\qquad\text{ Also written }\quad
&&\varnothing^\complement = X
&&\quad&&\text{ (Complement laws for the empty set))}
\\[1.4ex]

X \setminus X
&= \varnothing
&&\qquad\text{ Also written }\quad
&&X^\complement = \varnothing
&&\quad&&\text{ (Complement laws for the universe set)}
\\[1.4ex]
\end{alignat}</math>

==Two sets involved==

In the left hand sides of the following identities, <math>L</math> is the {{em|L}}{{hsp}}eft most set and <math>R</math> is the {{em|R}}{{hsp}}ight most set. 
Assume both <math>L \text{ and } R</math> are subsets of some universe set <math>X.</math>

===Formulas for binary set operations ⋂, ⋃, \, and ∆===

In the left hand sides of the following identities, <math>L</math> is the {{em|L}}{{hsp}}eft most set and <math>R</math> is the {{em|R}}{{hsp}}ight most set. Whenever necessary, both <math>L \text{ and } R</math> should be assumed to be subsets of some universe set <math>X,</math> so that <math>L^\complement := X \setminus L \text{ and } R^\complement := X \setminus R.</math>

<math display=block>\begin{alignat}{9}
L \cap R
&= L &&\,\,\setminus\,   &&(L &&\,\,\setminus &&R) \\
&= R &&\,\,\setminus\,   &&(R &&\,\,\setminus &&L) \\
&= L &&\,\,\setminus\,   &&(L &&\,\triangle\, &&R) \\
&= L &&\,\triangle\,     &&(L &&\,\,\setminus &&R) \\
\end{alignat}</math>

<math display=block>\begin{alignat}{9}
L \cup R
&= (&&L \,\triangle\, R) &&\,\,\cup      && &&L &&       &&  \\
&= (&&L \,\triangle\, R) &&\,\triangle\, &&(&&L &&\cap\, &&R) \\
&= (&&R \,\setminus\, L) &&\,\,\cup      && &&L &&       &&    ~~~~~\text{ (union is disjoint)} \\
\end{alignat}</math>

<math display=block>\begin{alignat}{9}
L \,\triangle\, R
&=  &&R \,\triangle\, L    &&                  && &&                     && \\
&= (&&L \,\cup\, R)        &&\,\setminus\,     &&(&&L \,\,\cap\, R)      && \\
&= (&&L \,\setminus\, R)   &&\cup\,            &&(&&R \,\,\setminus\, L) && ~~~~~\text{ (union is disjoint)} \\
&= (&&L \,\triangle\, M)   &&\,\triangle\,     &&(&&M \,\triangle\, R)   && ~~~~~\text{ where } M \text{ is an arbitrary set. } \\
&= (&&L^\complement)                 &&\,\triangle\,     &&(&&R^\complement)                 && \\
\end{alignat}</math>

<math display=block>\begin{alignat}{9}
L \setminus R
&=   &&L                   &&\,\,\setminus     &&(L &&\,\,\cap       &&R) \\
&=   &&L                   &&\,\,\cap          &&(L &&\,\triangle\,  &&R) \\
&=   &&L                   &&\,\triangle\,     &&(L &&\,\,\cap       &&R) \\
&=   &&R                   &&\,\triangle\,     &&(L &&\,\,\cup       &&R) \\
\end{alignat}</math>

===De Morgan's laws===

'''[[De Morgan's laws]]''' state that for <math>L, R \subseteq X:</math>

<math display=block>\begin{alignat}{10}
X \setminus (L \cap R)
&= (X \setminus L) \cup (X \setminus R)
&&\qquad\text{ Also written }\quad
&&(L \cap R)^\complement = L^\complement \cup R^\complement
&&\quad&&\text{ (De Morgan's law)}
\\[1.4ex]

X \setminus (L \cup R)
&= (X \setminus L) \cap (X \setminus R)
&&\qquad\text{ Also written }\quad
&&(L \cup R)^\complement = L^\complement \cap R^\complement
&&\quad&&\text{ (De Morgan's law)}
\\[1.4ex]
\end{alignat}</math>

===Commutativity===

Unions, intersection, and symmetric difference are '''commutative operations''':{{sfn|Monk|1969|pp=24-54}}

<math display=block>\begin{alignat}{10}
L \cup R                &\;=\;&& R \cup L        && \quad \text{ (Commutativity)} \\[1.4ex]
L \cap R                &\;=\;&& R \cap L        && \quad \text{ (Commutativity)} \\[1.4ex]
L \,\triangle R         &\;=\;&& R \,\triangle L && \quad \text{ (Commutativity)} \\[1.4ex]
\end{alignat}</math>

Set subtraction is not commutative. However, the commutativity of set subtraction can be characterized: from <math>(L \,\setminus\, R) \cap (R \,\setminus\, L) = \varnothing</math> it follows that:
<math display=block>L \,\setminus\, R = R \,\setminus\, L \quad \text{ if and only if } \quad L = R.</math> 
Said differently, if distinct symbols always represented distinct sets, then the {{em|only}} true formulas of the form <math>\,\cdot\, \,\setminus\, \,\cdot\, = \,\cdot\, \,\setminus\, \,\cdot\,</math> that could be written would be those involving a single symbol; that is, those of the form: <math>S \,\setminus\, S = S \,\setminus\, S.</math> 
But such formulas are necessarily true for {{em|every}} binary operation <math>\,\ast\,</math> (because <math>x \,\ast\, x = x \,\ast\, x</math> must hold by definition of [[Equality (mathematics)|equality]]), and so in this sense, set subtraction is as diametrically opposite to being commutative as is possible for a binary operation. 
Set subtraction is also neither left alternative nor right alternative; instead, <math>(L \setminus L) \setminus R = L \setminus (L \setminus R)</math> if and only if <math>L \cap R = \varnothing</math> if and only if <math>(R \setminus L) \setminus L = R \setminus (L \setminus L).</math> 
Set subtraction is quasi-commutative and satisfies the Jordan identity.

===Other identities involving two sets===

'''[[Absorption law]]s''':

<math display=block>\begin{alignat}{4}
L \cup (L \cap R)  &\;=\;&& L  && \quad \text{ (Absorption)} \\[1.4ex]
L \cap (L \cup R)  &\;=\;&& L  && \quad \text{ (Absorption)} \\[1.4ex]
\end{alignat}</math>

'''Other properties'''

<math display=block>\begin{alignat}{10}
L \setminus R
&= L \cap (X \setminus R)
&&\qquad\text{ Also written }\quad
&&L \setminus R = L \cap R^\complement
&&\quad&&\text{ where } L, R \subseteq X
\\[1.4ex]

X \setminus (L \setminus R)
&= (X \setminus L) \cup R
&&\qquad\text{ Also written }\quad
&&(L \setminus R)^\complement = L^\complement \cup R
&&\quad&&\text{ where } R \subseteq X
\\[1.4ex]

L \setminus R
&= (X \setminus R) \setminus (X \setminus L)
&&\qquad\text{ Also written }\quad
&&L \setminus R = R^\complement \setminus L^\complement
&&\quad&&\text{ where } L, R \subseteq X
\\[1.4ex]
\end{alignat}</math>

'''Intervals''':

<math display=block>(a, b) \cap (c, d) = (\max\{a,c\}, \min\{b,d\})</math>
<math display=block>[a, b) \cap [c, d) = [\max\{a,c\}, \min\{b,d\})</math>

===Subsets ⊆ and supersets ⊇===

The following statements are equivalent for any <math>L, R \subseteq X:</math>{{sfn|Monk|1969|pp=24-54}}
<ol>
<li><math>L \subseteq R</math>
* Definition of {{em|[[Subset|subset]]}}: if <math>l \in L</math> then <math>l \in R</math></li>
<li></li>
<li><math>L \cap R = L</math></li>
<li><math>L \cup R = R</math></li>
<li><math>L \,\triangle\, R = R \setminus L</math></li>
<li><math>L \,\triangle\, R \subseteq R \setminus L</math></li>
<li><math>L \setminus R = \varnothing</math></li>
<li><math>L</math> and <math>X \setminus R</math> are disjoint (that is, <math>L \cap (X \setminus R) = \varnothing</math>)</li>
<li><math>X \setminus R \subseteq X \setminus L \qquad</math> (that is, <math>R^\complement \subseteq L^\complement</math>)</li>
</ol>

The following statements are equivalent for any <math>L, R \subseteq X:</math>
<ol>
<li><math>L \not\subseteq R</math></li>
<li>There exists some <math>l \in L \setminus R.</math></li>
</ol>

====Set equality====

The following statements are equivalent:
<ol>
<li><math>L = R</math></li>
<li><math>L \,\triangle\, R = \varnothing</math></li>
<li><math>L \,\setminus\, R = R \,\setminus\, L</math></li>
</ol>

<ul>
<li>If <math>L \cap R = \varnothing</math> then <math>L = R</math> if and only if <math>L = \varnothing = R.</math></li>
<li>'''Uniqueness of complements''': If <math display=inline>L \cup R = X \text{ and } L \cap R = \varnothing</math> then <math>R = X \setminus L</math></li>
</ul>

=====Empty set=====

A set <math>L</math> is '''[[Axiom of empty set|empty]]''' if the sentence <math>\forall x (x \not\in L)</math> is true, where the notation <math>x \not\in L</math> is shorthand for <math>\lnot (x \in L).</math> 

If <math>L</math> is any set then the following are equivalent:
<ol>
<li><math>L</math> is not empty, meaning that the sentence <math>\lnot [\forall x (x \not\in L)]</math> is true (literally, the logical negation of "<math>L</math> is empty" holds true).</li>
<li>(In [[Classical mathematics|classical mathematics]]) <math>L</math> is [[Inhabited set|inhabited]], meaning: <math>\exists x (x \in L)</math>
* In constructive mathematics, "not empty" and "inhabited" are not equivalent: every inhabited set is not empty but the converse is not always guaranteed; that is, in constructive mathematics, a set <math>L</math> that is not empty (where by definition, "<math>L</math> is empty" means that the statement <math>\forall x (x \not\in L)</math> is true) might not have an inhabitant (which is an <math>x</math> such that <math>x \in L</math>).
</li>
<li><math>L \not\subseteq R</math> for some set <math>R</math></li>
</ol>

If <math>L</math> is any set then the following are equivalent:
<ol>
<li><math>L</math> is empty (<math>L = \varnothing</math>), meaning: <math>\forall x (x \not\in L)</math></li>
<li><math>L \cup R \subseteq R</math> for every set <math>R</math></li>
<li><math>L \subseteq R</math> for every set <math>R</math></li>
<li><math>L \subseteq R \setminus L</math> for some/every set <math>R</math></li>
<li><math>\varnothing / L = L</math></li>
</ol>

Given any <math>x,</math> <math display=block>x \not\in L \setminus R \quad \text{ if and only if }\quad  x \in L \cap R \; \text{ or } \; x \not\in L.</math>

Moreover,
<math display=block>(L \setminus R) \cap R = \varnothing \qquad \text{ always holds}.</math>

====Meets, Joins, and lattice properties====

'''Inclusion is a partial order''': 
Explicitly, this means that [[Subset|inclusion]] <math>\,\subseteq,\,</math> which is a [[Binary operation|binary operation]], has the following three properties:{{sfn|Monk|1969|pp=24-54}}
<ul>
<li>'''[[Reflexive relation|Reflexivity]]''': 
<math display=inline>L \subseteq L</math></li>
<li>'''[[Antisymmetric relation|Antisymmetry]]''': 
<math display=inline>(L \subseteq R \text{ and } R \subseteq L) \text{ if and only if } L = R</math></li>
<li>'''[[Transitive relation|Transitivity]]''': 
<math display=inline>\text{If }L \subseteq M \text{ and } M \subseteq R \text{ then } L \subseteq R</math></li>
</ul>

The following proposition says that for any set <math>S,</math> the [[Power set|power set]] of <math>S,</math> ordered by inclusion, is a [[Lattice (order)|bounded lattice]], and hence together with the distributive and complement laws above, show that it is a [[Boolean algebra (structure)|Boolean algebra]].

'''Existence of a least element and a greatest element''': 
<math display=block>\varnothing \subseteq L \subseteq X</math>

'''Joins/supremums exist''':{{sfn|Monk|1969|pp=24-54}} 
<math display=block>L \subseteq L \cup R</math>

The union <math>L \cup R</math> is the join/supremum of <math>L</math> and <math>R</math> with respect to <math>\,\subseteq\,</math> because:
<ol>
<li><math>L \subseteq L \cup R</math> and <math>R \subseteq L \cup R,</math> and</li>
<li>if <math>Z</math> is a set such that <math>L \subseteq Z</math> and <math>R \subseteq Z</math> then <math>L \cup R \subseteq Z.</math></li>
</ol>

The intersection <math>L \cap R</math> is the join/supremum of <math>L</math> and <math>R</math> with respect to <math>\,\supseteq.\,</math>

'''Meets/infimums exist''':{{sfn|Monk|1969|pp=24-54}} 
<math display=block>L \cap R \subseteq L</math>

The intersection <math>L \cap R</math> is the meet/infimum of <math>L</math> and <math>R</math> with respect to <math>\,\subseteq\,</math> because:
<ol>
<li>if <math>L \cap R \subseteq L</math> and <math>L \cap R \subseteq R,</math> and</li>
<li>if <math>Z</math> is a set such that <math>Z \subseteq L</math> and <math>Z \subseteq R</math> then <math>Z \subseteq L \cap R.</math></li>
</ol>

The union <math>L \cup R</math> is the meet/infimum of <math>L</math> and <math>R</math> with respect to <math>\,\supseteq.\,</math>

'''Other inclusion properties''':

<math display=block>L \setminus R \subseteq L</math>
<math display=block>(L \setminus R) \cap L = L \setminus R</math>

<ul>
<li>If <math>L \subseteq R</math> then <math>L \,\triangle\, R = R \setminus L.</math></li>
<li>If <math>L \subseteq X</math> and <math>R \subseteq Y</math> then <math>L \times R \subseteq X \times Y</math>{{sfn|Monk|1969|pp=24-54}}</li>
</ul>

==Three sets involved==

In the left hand sides of the following identities, <math>L</math> is the {{em|L}}{{hsp}}eft most set, <math>M</math> is the {{em|M}}{{hsp}}iddle set, and <math>R</math> is the {{em|R}}{{hsp}}ight most set. 

'''Precedence rules'''

There is no universal agreement on the order of precedence of the basic set operators.
Nevertheless, many authors use precedence rules for set operators, although these rules vary with the author. 

One common convention is to associate intersection <math>L \cap R = \{x : (x \in L) \land (x \in R)\}</math> with [[Logical conjunction|logical conjunction (and)]] <math>L \land R</math> and associate union <math>L \cup R = \{x : (x \in L) \lor (x \in R)\}</math> with [[Logical disjunction|logical disjunction (or)]] <math>L \lor R,</math> and then transfer the [[Logical connective#Order of precedence|precedence of these logical operator]]s (where <math>\,\land\,</math> has precedence over <math>\,\lor\,</math>) to these set operators, thereby giving <math>\,\cap\,</math> precedence over <math>\,\cup.\,</math> 
So for example, <math>L \cup M \cap R</math> would mean <math>L \cup (M \cap R)</math> since it would be associated with the logical statement <math>L \lor M \land R ~=~ L \lor (M \land R)</math> and similarly, <math>L \cup M \cap R \cup Z</math> would mean <math>L \cup (M \cap R) \cup Z</math> since it would be associated with <math>L \lor M \land R \lor Z ~=~ L \lor (M \land R) \lor Z.</math> 

Sometimes, set complement (subtraction) <math>\,\setminus\,</math> is also associated with logical complement (not) <math>\,\lnot,\,</math> in which case it will have the highest precedence. 
More specifically, <math>L \setminus R = \{x : (x \in L) \land \lnot (x \in R)\}</math> is rewritten <math>L \land \lnot R</math> so that for example, <math>L \cup M \setminus R</math> would mean <math>L \cup (M \setminus R)</math> since it would be rewritten as the logical statement <math>L \lor M \land \lnot R</math> which is equal to <math>L \lor (M \land \lnot R).</math>
For another example, because <math>L \land \lnot M \land R</math> means <math>L \land (\lnot M) \land R,</math> which is equal to both <math>(L \land (\lnot M)) \land R</math> and <math>L \land ((\lnot M) \land R) ~=~ L \land (R \land (\lnot M))</math> (where <math>(\lnot M) \land R</math> was rewritten as <math>R \land (\lnot M)</math>), the formula <math>L \setminus M \cap R</math> would refer to the set <math>(L \setminus M) \cap R = L \cap (R \setminus M);</math> 
moreover, since <math>L \land (\lnot M) \land R = (L \land R) \land \lnot M,</math> this set is also equal to <math>(L \cap R) \setminus M</math> (other set identities can similarly be deduced from [[Propositional calculus|propositional calculus]] [[Logical equivalence|identities]] in this way).
However, because set subtraction is not associative <math>(L \setminus M) \setminus R \neq L \setminus (M \setminus R),</math> a formula such as <math>L \setminus M \setminus R</math> would be ambiguous; for this reason, among others, set subtraction is often not assigned any precedence at all. 

[[Symmetric difference]] <math>L \triangle R = \{x : (x \in L) \oplus (x \in R)\}</math> is sometimes associated with [[Exclusive or|exclusive or (xor)]] <math>L \oplus R</math> (also sometimes denoted by <math>\,\veebar</math>), in which case if the order of precedence from highest to lowest is <math>\,\lnot, \,\oplus, \,\land, \,\lor\,</math> then the order of precedence (from highest to lowest) for the set operators would be <math>\,\setminus,\, \triangle,\, \cap,\, \cup.</math> 
There is no universal agreement on the precedence of exclusive disjunction <math>\,\oplus\,</math> with respect to the other logical connectives, which is why symmetric difference <math>\,\triangle\,</math> is not often assigned a precedence.

===Associativity===

'''Definition''': A binary operator <math>\,\ast\,</math> is called '''associative''' if <math>(L \,\ast\, M) \,\ast\, R = L \,\ast\, (M \,\ast\, R)</math> always holds. 

The following set operators are associative:{{sfn|Monk|1969|pp=24-54}}

<math display=block>\begin{alignat}{5}
(L \cup M) \cup R               &\;=\;\;&& L \cup (M \cup R) \\[1.4ex]
(L \cap M) \cap R               &\;=\;\;&& L \cap (M \cap R) \\[1.4ex]
(L \,\triangle M) \,\triangle R &\;=\;\;&& L \,\triangle (M \,\triangle R) \\[1.4ex]
\end{alignat}</math>

For set subtraction, instead of associativity, only the following is always guaranteed:
<math display=block>(L \,\setminus\, M) \,\setminus\, R \;~~{\color{red}{\subseteq}}~~\; L \,\setminus\, (M \,\setminus\, R)</math>
where equality holds if and only if <math>L \cap R = \varnothing</math> (this condition does not depend on <math>M</math>). Thus 
<math display=inline>\; (L \setminus M) \setminus R = L \setminus (M \setminus R) \;</math> if and only if <math>\; (R \setminus M) \setminus L = R \setminus (M \setminus L), \;</math> 
where the only difference between the left and right hand side set equalities is that the locations of <math>L \text{ and } R</math> have been swapped.

===Distributivity===

'''Definition''': If <math>\ast \text{ and } \bullet</math> are binary operators then {{em|<math>\,\ast\,</math> '''left distributes''' over <math>\,\bullet\,</math>}} if <math display=block>L \,\ast\, (M \,\bullet\, R) ~=~ (L \,\ast\, M) \,\bullet\, (L \,\ast\,R) \qquad\qquad \text{ for all } L, M, R</math>
while {{em|<math>\,\ast\,</math> '''right distributes''' over <math>\,\bullet\,</math>}} if <math display=block>(L \,\bullet\, M) \,\ast\, R ~=~ (L \,\ast\, R) \,\bullet\, (M \,\ast\,R) \qquad\qquad \text{ for all } L, M, R.</math> 
The operator {{em|<math>\,\ast\,</math> '''distributes''' over <math>\,\bullet\,</math>}} if it both left distributes and right distributes over <math>\,\bullet\,.\,</math> 
In the definitions above, to transform one side to the other, the innermost operator (the operator inside the parentheses) becomes the outermost operator and the outermost operator becomes the innermost operator.

'''Right distributivity''':{{sfn|Monk|1969|pp=24-54}}

<math display=block>\begin{alignat}{9}
(L \,\cap\, M) \,\cup\, R      ~&~~=~~&& (L \,\cup\, R) \,&&\cap\, &&(M \,\cup\, R)             \qquad 
&&\text{ (Right-distributivity of } \,\cup\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
(L \,\cup\, M) \,\cup\, R      ~&~~=~~&& (L \,\cup\, R) \,&&\cup\, &&(M \,\cup\, R)             \qquad 
&&\text{ (Right-distributivity of } \,\cup\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
(L \,\cup\, M) \,\cap\, R      ~&~~=~~&& (L \,\cap\, R) \,&&\cup\, &&(M \,\cap\, R)             \qquad 
&&\text{ (Right-distributivity of } \,\cap\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
(L \,\cap\, M) \,\cap\, R      ~&~~=~~&& (L \,\cap\, R) \,&&\cap\, &&(M \,\cap\, R)             \qquad 
&&\text{ (Right-distributivity of } \,\cap\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
(L \,\triangle\, M) \,\cap\, R      ~&~~=~~&& (L \,\cap\, R) \,&&\triangle\, &&(M \,\cap\, R)    \qquad 
&&\text{ (Right-distributivity of } \,\cap\, \text{ over } \,\triangle\, \text{)} \\[1.4ex]
(L \,\cap\, M) \,\times\, R      ~&~~=~~&& (L \,\times\, R) \,&&\cap\, &&(M \,\times\, R)             \qquad 
&&\text{ (Right-distributivity of } \,\times\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
(L \,\cup\, M) \,\times\, R      ~&~~=~~&& (L \,\times\, R) \,&&\cup\, &&(M \,\times\, R)              \qquad 
&&\text{ (Right-distributivity of } \,\times\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
(L \,\setminus\, M) \,\times\, R      ~&~~=~~&& (L \,\times\, R) \,&&\setminus\, &&(M \,\times\, R)    \qquad 
&&\text{ (Right-distributivity of } \,\times\, \text{ over } \,\setminus\, \text{)} \\[1.4ex]
(L \,\cup\, M) \,\setminus\, R      ~&~~=~~&& (L \,\setminus\, R) \,&&\cup\, &&(M \,\setminus\, R)      \qquad 
&&\text{ (Right-distributivity of } \,\setminus\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
(L \,\cap\, M) \,\setminus\, R      ~&~~=~~&& (L \,\setminus\, R) \,&&\cap\, &&(M \,\setminus\, R)      \qquad 
&&\text{ (Right-distributivity of } \,\setminus\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
(L \,\triangle\, M) \,\setminus\, R  ~&~~=~~&& (L \,\setminus\, R) &&\,\triangle\, &&(M \,\setminus\, R) \qquad 
&&\text{ (Right-distributivity of } \,\setminus\, \text{ over } \,\triangle\, \text{)} \\[1.4ex]
(L \,\setminus\, M) \,\setminus\, R  ~&~~=~~&& (L \,\setminus\, R) &&\,\setminus\, &&(M \,\setminus\, R)  \qquad 
&&\text{ (Right-distributivity of } \,\setminus\, \text{ over } \,\setminus\, \text{)} \\[1.4ex]
~&~~=~~&&~~\;~~\;~~\;~ L &&\,\setminus\, &&(M \cup R) \\[1.4ex]
\end{alignat}</math>

'''Left distributivity''':{{sfn|Monk|1969|pp=24-54}}

<math display=block>\begin{alignat}{5}
L \cup   (M \cap R)            &\;=\;\;&& (L \cup M)     \cap        (L \cup R) \qquad
&&\text{ (Left-distributivity of } \,\cup\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
L \cup   (M \cup R)            &\;=\;\;&& (L \cup M)     \cup        (L \cup R) 
&&\text{ (Left-distributivity of } \,\cup\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
L \cap   (M \cup R)            &\;=\;\;&& (L \cap M)     \cup        (L \cap R) 
&&\text{ (Left-distributivity of } \,\cap\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
L \cap   (M \cap R)            &\;=\;\;&& (L \cap M)     \cap        (L \cap R) 
&&\text{ (Left-distributivity of } \,\cap\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
L \cap   (M \,\triangle\, R)   &\;=\;\;&& (L \cap M)   \,\triangle\, (L \cap R) 
&&\text{ (Left-distributivity of } \,\cap\, \text{ over } \,\triangle\, \text{)} \\[1.4ex]
L \times (M \cap R)            &\;=\;\;&& (L \times M)   \cap        (L \times R) 
&&\text{ (Left-distributivity of } \,\times\, \text{ over } \,\cap\, \text{)} \\[1.4ex]
L \times (M \cup R)            &\;=\;\;&& (L \times M)   \cup        (L \times R) 
&&\text{ (Left-distributivity of } \,\times\, \text{ over } \,\cup\, \text{)} \\[1.4ex]
L \times (M \,\setminus R)     &\;=\;\;&& (L \times M) \,\setminus   (L \times R) 
&&\text{ (Left-distributivity of } \,\times\, \text{ over } \,\setminus\, \text{)} \\[1.4ex]
\end{alignat}</math>

====Distributivity and symmetric difference ∆====

Intersection distributes over symmetric difference:
<math display=block>\begin{alignat}{5}
L \,\cap\, (M \,\triangle\, R) ~&~~=~~&&  (L \,\cap\, M) \,\triangle\, (L \,\cap\, R) ~&&~ \\[1.4ex]
\end{alignat}</math>
<math display=block>\begin{alignat}{5}
(L \,\triangle\, M) \,\cap\, R~&~~=~~&&  (L \,\cap\, R) \,\triangle\, (M \,\cap\, R) ~&&~ \\[1.4ex]
\end{alignat}</math>

Union does not distribute over symmetric difference because only the following is guaranteed in general:
<math display=block>\begin{alignat}{5}
L \cup (M \,\triangle\, R) 
~~{\color{red}{\supseteq}}~~ \color{black}{\,} (L \cup M) \,\triangle\, (L \cup R) ~
&~=~&& (M \,\triangle\, R) \,\setminus\, L  
&~=~&& (M \,\setminus\, L) \,\triangle\, (R \,\setminus\, L) \\[1.4ex]
\end{alignat}</math>

Symmetric difference does not distribute over itself:
<math display=block>L \,\triangle\, (M \,\triangle\, R) 
~~{\color{red}{\neq}}~~ \color{black}{\,} (L \,\triangle\, M) \,\triangle\, (L \,\triangle\, R) 
~=~ M \,\triangle\, R</math>
and in general, for any sets <math>L \text{ and } A</math> (where <math>A</math> represents <math>M \,\triangle\, R</math>), <math>L \,\triangle\, A</math> might not be a subset, nor a superset, of <math>L</math> (and the same is true for <math>A</math>).

====Distributivity and set subtraction \====

'''Failure of set subtraction to left distribute''':

Set subtraction is {{em|right}} distributive over itself. However, set subtraction is {{em|not}} left distributive over itself because only the following is guaranteed in general:
<math display=block>\begin{alignat}{5}
L \,\setminus\, (M \,\setminus\, R) &~~{\color{red}{\supseteq}}~~&& \color{black}{\,} (L \,\setminus\, M) \,\setminus\, (L \,\setminus\, R) ~~=~~ L \cap R \,\setminus\, M \\[1.4ex]
\end{alignat}</math>
where equality holds if and only if <math>L \,\setminus\, M = L \,\cap\, R,</math> which happens if and only if <math>L \cap M \cap R = \varnothing \text{ and } L \setminus M \subseteq R.</math> 

For symmetric difference, the sets <math>L \,\setminus\, (M \,\triangle\, R)</math> and <math>(L \,\setminus\, M) \,\triangle\, (L \,\setminus\, R) = L \,\cap\, (M \,\triangle\, R)</math> are always disjoint. 
So these two sets are equal if and only if they are both equal to <math>\varnothing.</math> 
Moreover, <math>L \,\setminus\, (M \,\triangle\, R) = \varnothing</math> if and only if <math>L \cap M \cap R = \varnothing \text{ and } L \subseteq M \cup R.</math> 

To investigate the left distributivity of set subtraction over unions or intersections, consider how the sets involved in (both of) De Morgan's laws are all related:
<math display=block>\begin{alignat}{5}
(L \,\setminus\, M) \,\cap\, (L \,\setminus\, R) ~~=~~ L \,\setminus\, (M \,\cup\, R) ~&~~{\color{red}{\subseteq}}~~&& \color{black}{\,} L \,\setminus\, (M \,\cap\, R) ~~=~~ (L \,\setminus\, M) \,\cup\, (L \,\setminus\, R) \\[1.4ex]
\end{alignat}</math>
always holds (the equalities on the left and right are De Morgan's laws) but equality is not guaranteed in general (that is, the containment <math>{\color{red}{\subseteq}}</math> might be strict). 
Equality holds if and only if <math>L \,\setminus\, (M \,\cap\, R) \;\subseteq\; L \,\setminus\, (M \,\cup\, R),</math> which happens if and only if <math>L \,\cap\, M = L \,\cap\, R.</math> 

This observation about De Morgan's laws shows that <math>\,\setminus\,</math> is {{em|not}} left distributive over <math>\,\cup\,</math> or <math>\,\cap\,</math> because only the following are guaranteed in general:
<math display=block>\begin{alignat}{5}
L \,\setminus\, (M \,\cup\, R) ~&~~{\color{red}{\subseteq}}~~&& \color{black}{\,} (L \,\setminus\, M) \,\cup\, (L \,\setminus\, R) ~~=~~ L \,\setminus\, (M \,\cap\, R) \\[1.4ex]
\end{alignat}</math>
<math display=block>\begin{alignat}{5}
L \,\setminus\, (M \,\cap\, R) ~&~~{\color{red}{\supseteq}}~~&& \color{black}{\,} (L \,\setminus\, M) \,\cap\, (L \,\setminus\, R) ~~=~~ L \,\setminus\, (M \,\cup\, R) \\[1.4ex]
\end{alignat}</math>
where equality holds for one (or equivalently, for both) of the above two inclusion formulas if and only if <math>L \,\cap\, M = L \,\cap\, R.</math> 

The following statements are equivalent:
<ol>
<li><math>L \cap M \,=\, L \cap R</math></li>
<li><math>L \,\setminus\, M \,=\, L \,\setminus\, R</math></li>
<li><math>L \,\setminus\, (M \,\cap\, R) = (L \,\setminus\, M) \,\cap\, (L \,\setminus\, R);</math> that is, <math>\,\setminus\,</math> left distributes over <math>\,\cap\,</math> for these three particular sets</li>
<li><math>L \,\setminus\, (M \,\cup\, R) = (L \,\setminus\, M) \,\cup\, (L \,\setminus\, R);</math> that is, <math>\,\setminus\,</math> left distributes over <math>\,\cup\,</math> for these three particular sets</li>
<li><math>L \,\setminus\, (M \,\cap\, R) \,=\, L \,\setminus\, (M \,\cup\, R)</math></li>
<li><math>L \cap (M \cup R) \,=\, L \cap M \cap R</math></li>
<li><math>L \cap (M \cup R) ~\subseteq~ M \cap R</math></li>
<li><math>L \cap R ~\subseteq~ M \;</math> and <math>\; L \cap M ~\subseteq~ R</math></li>
<li><math>L \setminus (M \setminus R) \,=\, L \setminus (R \setminus M)</math></li>
<li><math>L \setminus (M \setminus R) \,=\, L \setminus (R \setminus M) \,=\, L</math></li>
</ol>

'''[[Quasi-commutative property|Quasi-commutativity]]''': 
<math display=block>(L \setminus M) \setminus R ~=~ (L \setminus R) \setminus M \qquad \text{ (Quasi-commutative)}</math>
always holds but in general, 
<math display=block>L \setminus (M \setminus R) ~~{\color{red}{\neq}}~~ L \setminus (R \setminus M).</math>
However, <math>L \setminus (M \setminus R) ~\subseteq~ L \setminus (R \setminus M)</math> if and only if <math>L \cap R ~\subseteq~ M</math> if and only if <math>L \setminus (R \setminus M) ~=~ L.</math> 

'''Set subtraction complexity''': To manage the many identities involving set subtraction, this section is divided based on where the set subtraction operation and parentheses are located on the left hand side of the identity. The great variety and (relative) complexity of formulas involving set subtraction (compared to those without it) is in part due to the fact that unlike <math>\,\cup, \,\cap,</math> and <math>\triangle,\,</math> set subtraction is neither associative nor commutative and it also is not left distributive over <math>\,\cup, \,\cap, \,\triangle,</math> or even over itself.

===Two set subtractions===

Set subtraction is {{em|not}} associative in general: 
<math display=block>(L \,\setminus\, M) \,\setminus\, R \;~~{\color{red}{\neq}}~~\; L \,\setminus\, (M \,\setminus\, R)</math>
since only the following is always guaranteed:
<math display=block>(L \,\setminus\, M) \,\setminus\, R \;~~{\color{red}{\subseteq}}~~\; L \,\setminus\, (M \,\setminus\, R).</math>

====(L\M)\R ====

<math display=block>\begin{alignat}{4}
(L \setminus M) \setminus R 
&=  &&L \setminus (M \cup R) \\[0.6ex]
&= (&&L \setminus R) \setminus M \\[0.6ex]
&= (&&L \setminus M) \cap (L \setminus R) \\[0.6ex]
&= (&&L \setminus R) \setminus M \\[0.6ex]
&= (&&L \,\setminus\, R) \,\setminus\, (M \,\setminus\, R) \\[1.4ex]
\end{alignat}</math>

====L\(M\R) ====

<math display=block>\begin{alignat}{4}
L \setminus (M \setminus R) 
&= (L \setminus M) \cup (L \cap R) \\[1.4ex]
\end{alignat}</math>
:* If <math>L \subseteq M \text{ then } L \setminus (M \setminus R) = L \cap R</math>
:* <math display=inline>L \setminus (M \setminus R) \subseteq (L \setminus M) \cup R</math> with equality if and only if <math>R \subseteq L.</math>

===One set subtraction===

====(L\M) ⁎ R ====

'''Set subtraction on the ''left'', and parentheses on the {{em|left}}'''

<math display=block>\begin{alignat}{4}
\left(L \setminus M\right) \cup R 
&= (L \cup R) \setminus (M \setminus R) \\
&= (L \setminus (M \cup R)) \cup R ~~~~~ \text{ (the outermost union is disjoint) } \\
\end{alignat}</math>

:<math>\begin{alignat}{4}
(L \setminus M) \cap R 
&= (&&L \cap R) \setminus (M \cap R) ~~~\text{ (Distributive law of } \cap \text{ over } \setminus \text{ )} \\
&= (&&L \cap R) \setminus M \\
&=  &&L \cap (R \setminus M) \\
\end{alignat}</math>{{sfn|Császár|1978|pp=15-26}}

<math display=block>\begin{alignat}{5}
(L \,\setminus\, M) \,\cap\, (L \,\setminus\, R) ~~=~~ L \,\setminus\, (M \,\cup\, R) ~&~~{\color{red}{\subseteq}}~~&& \color{black}{\,} L \,\setminus\, (M \,\cap\, R) ~~=~~ (L \,\setminus\, M) \,\cup\, (L \,\setminus\, R) \\[1.4ex]
\end{alignat}</math> 
<math display=block>\begin{alignat}{4}
(L \setminus M) ~\triangle~ R 
&= (L \setminus (M \cup R)) \cup (R \setminus L) \cup (L \cap M \cap  R) ~~~\text{ (the three outermost sets are pairwise disjoint) } \\
\end{alignat}</math>

<math display=block>(L \,\setminus M) \times R = (L \times R) \,\setminus (M \times R)    ~~~~~\text{ (Distributivity)}</math>

====L\(M ⁎ R) ====

'''Set subtraction on the ''left'', and parentheses on the {{em|right}}'''

<math display=block>\begin{alignat}{3}
L \setminus (M \cup R) 
&= (L \setminus M) &&\,\cap\,    (&&L \setminus R) ~~~~\text{ (De Morgan's law) } \\
&= (L \setminus M) &&\,\,\setminus &&R \\
&= (L \setminus R) &&\,\,\setminus &&M \\
\end{alignat}</math>

<math display=block>\begin{alignat}{4}
L \setminus (M \cap R) 
&= (L \setminus M) \cup (L \setminus R) ~~~~\text{ (De Morgan's law) } \\
\end{alignat}</math>
where the above two sets that are the subjects of [[De Morgan's laws]] always satisfy <math>L \,\setminus\, (M \,\cup\, R) ~~{\color{red}{\subseteq}}~~ \color{black}{\,} L \,\setminus\, (M \,\cap\, R).</math>

<math display=block>\begin{alignat}{4}
L \setminus (M ~\triangle~ R) 
&= (L \setminus (M \cup R)) \cup (L \cap M \cap R) ~~~\text{ (the outermost union is disjoint) } \\
\end{alignat}</math>

====(L ⁎ M)\R ====

'''Set subtraction on the ''right'', and parentheses on the {{em|left}}'''

<math display=block>\begin{alignat}{4}
(L \cup M) \setminus R 
&= (L \setminus R) \cup (M \setminus R) \\
\end{alignat}</math>

<math display=block>\begin{alignat}{4}
(L \cap M) \setminus R 
&= (&&L \setminus R) &&\cap (M \setminus R) \\
&=  &&L              &&\cap (M \setminus R) \\
&=  &&M              &&\cap (L \setminus R) \\
\end{alignat}</math>

<math display=block>\begin{alignat}{4}
(L \,\triangle\, M)   \setminus R 
&= (L \setminus R) ~&&\triangle~ (M \setminus R) \\
&= (L \cup R)      ~&&\triangle~ (M \cup R) \\
\end{alignat}</math>

====L ⁎ (M\R) ====

'''Set subtraction on the ''right'', and parentheses on the {{em|right}}'''

<math display=block>\begin{alignat}{3}
L \cup (M \setminus R)
&=  && &&L              &&\cup\;            &&(M \setminus (R \cup L)) &&~~~\text{ (the outermost union is disjoint) } \\
&= [&&(&&L \setminus M) &&\cup\;            &&(R \cap L)] \cup (M \setminus R)       &&~~~\text{ (the outermost union is disjoint) } \\
&=  &&(&&L \setminus (M \cup R)) \;&&\;\cup &&(R \cap L)\,\, \cup (M \setminus R)   &&~~~\text{ (the three outermost sets are pairwise disjoint) } \\
\end{alignat}</math>

:<math>\begin{alignat}{4}
L \cap (M \setminus R)
&= (&&L \cap M)      &&\setminus (L \cap R) ~~~\text{ (Distributive law of } \cap \text{ over } \setminus \text{ )}  \\
&= (&&L \cap M)      &&\setminus R  \\
&=  &&M              &&\cap (L \setminus R) \\
&= (&&L \setminus R) &&\cap (M \setminus R) \\
\end{alignat}
</math>{{sfn|Császár|1978|pp=15-26}}

<math display=block>L \times (M \,\setminus R) = (L \times M) \,\setminus (L \times R)    ~~~~~\text{ (Distributivity)}</math>

===Three operations on three sets===

====(L • M) ⁎ (M • R) ====

'''Operations of the form <math>(L \bullet M) \ast (M \bullet R)</math>''':

<math display=block>\begin{alignat}{9}
(L \cup M)            &\,\cup\,&&       (&&M \cup R) && 
              &&\;=\;\;&& L \cup M \cup R \\[1.4ex]
(L \cup M)            &\,\cap\,&&       (&&M \cup R) && 
              &&\;=\;\;&& M \cup (L \cap R) \\[1.4ex]
(L \cup M)            &\,\setminus\,&&  (&&M \cup R) && 
              &&\;=\;\;&& L \,\setminus\, (M \cup R) \\[1.4ex]
(L \cup M)            &\,\triangle\,&&  (&&M \cup R) && 
              &&\;=\;\;&& (L \,\setminus\, (M \cup R)) \,\cup\, (R \,\setminus\, (L \cup M)) \\[1.4ex]
&\,&&\,&&\,&& &&\;=\;\;&& (L \,\triangle\, R) \,\setminus\, M \\[1.4ex]
(L \cap M)            &\,\cup\,&&       (&&M \cap R) && 
              &&\;=\;\;&& M \cup (L \cap R) \\[1.4ex]
(L \cap M)            &\,\cap\,&&       (&&M \cap R) && 
              &&\;=\;\;&& L \cap M \cap R \\[1.4ex]
(L \cap M)            &\,\setminus\,&&  (&&M \cap R) && 
              &&\;=\;\;&& (L \cap M) \,\setminus\, R \\[1.4ex]
(L \cap M)            &\,\triangle\,&&  (&&M \cap R) && 
              &&\;=\;\;&& [(L \,\cap M) \cup (M \,\cap R)] \,\setminus\, (L \,\cap M \,\cap R) \\[1.4ex]
(L \,\setminus M)     &\,\cup\,&&  (&&M \,\setminus R) && 
              &&\;=\;\;&& (L \,\cup M) \,\setminus (M \,\cap\, R) \\[1.4ex]
(L \,\setminus M)     &\,\cap\,&&  (&&M \,\setminus R) && 
              &&\;=\;\;&& \varnothing \\[1.4ex]
(L \,\setminus M)     &\,\setminus\,&&  (&&M \,\setminus R) && 
              &&\;=\;\;&& L \,\setminus M \\[1.4ex]
(L \,\setminus M)     &\,\triangle\,&&  (&&M \,\setminus R) && 
              &&\;=\;\;&& (L \,\setminus M) \cup (M \,\setminus R) \\[1.4ex]
&\,&&\,&&\,&& &&\;=\;\;&& (L \,\cup M) \setminus (M \,\cap R) \\[1.4ex]
(L \,\triangle\, M)   &\,\cup\,&&  (&&M \,\triangle\, R) && 
              &&\;=\;\;&& (L \,\cup\, M \,\cup\, R) \,\setminus\, (L \,\cap\, M \,\cap\, R) \\[1.4ex]
(L \,\triangle\, M)   &\,\cap\,&&  (&&M \,\triangle\, R) && 
              &&\;=\;\;&& ((L \,\cap\, R) \,\setminus\, M) \,\cup\, (M \,\setminus\, (L \,\cup\, R)) \\[1.4ex]
(L \,\triangle\, M)   &\,\setminus\,&&  (&&M \,\triangle\, R) && 
              &&\;=\;\;&& (L \,\setminus\, (M \,\cup\, R)) \,\cup\, ((M \,\cap\, R) \,\setminus\, L) \\[1.4ex]
(L \,\triangle\, M)   &\,\triangle\,&&  (&&M \,\triangle\, R) && 
              &&\;=\;\;&& L \,\triangle\, R \\[1.7ex]
\end{alignat}</math>

====(L • M) ⁎ (R\M) ====

'''Operations of the form <math>(L \bullet M) \ast (R \,\setminus\, M)</math>''':

<math display=block>\begin{alignat}{9}
(L \cup M)            &\,\cup\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& L \cup M \cup R \\[1.4ex]
(L \cup M)            &\,\cap\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& (L \cap R) \,\setminus\, M \\[1.4ex]
(L \cup M)            &\,\setminus\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& M \cup (L \,\setminus\, R) \\[1.4ex]
(L \cup M)            &\,\triangle\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& M \cup (L \,\triangle\, R) \\[1.4ex]
(L \cap M)            &\,\cup\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& [L \cap (M \cup R)] \cup [R \,\setminus\, (L \cup M)] \qquad \text{ (disjoint union)} \\[1.4ex]
&\,&&\,&&\,&& &&\;=\;\;&& (L \cap M) \,\triangle\, (R \,\setminus\, M) \\[1.4ex]
(L \cap M)            &\,\cap\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& \varnothing \\[1.4ex]
(L \cap M)            &\,\setminus\,&&   (&&R \,\setminus\, M) && 
              &&\;=\;\;&& L \cap M \\[1.4ex]
(L \cap M)            &\,\triangle\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& (L \cap M) \cup (R \,\setminus\, M) \qquad \text{ (disjoint union)} \\[1.4ex]
(L \,\setminus\, M)   &\,\cup\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& L \cup R \,\setminus\, M \\[1.4ex]
(L \,\setminus\, M)   &\,\cap\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& (L \cap R) \,\setminus\, M \\[1.4ex]
(L \,\setminus\, M)   &\,\setminus\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& L \,\setminus\, (M \cup R) \\[1.4ex]
(L \,\setminus\, M)   &\,\triangle\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& (L \,\triangle\, R) \,\setminus\, M \\[1.4ex]
(L \,\triangle\, M)   &\,\cup\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& (L \cup M \cup R) \,\setminus\, (L \cap M) \\[1.4ex]
(L \,\triangle\, M)   &\,\cap\,&&       (&&R \,\setminus\, M) && 
              &&\;=\;\;&& (L \cap R) \,\setminus\, M \\[1.4ex]
(L \,\triangle\, M)   &\,\setminus\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& [L \,\setminus\, (M \cup R)] \cup (M \,\setminus\, L) \qquad \text{ (disjoint union)} \\[1.4ex]
&\,&&\,&&\,&& &&\;=\;\;&& (L \,\triangle\, M) \setminus (L \,\cap R) \\[1.4ex]
(L \,\triangle\, M)   &\,\triangle\,&&  (&&R \,\setminus\, M) && 
              &&\;=\;\;&& L \,\triangle\, (M \cup R) \\[1.7ex]
\end{alignat}</math>

====(L\M) ⁎ (L\R) ====

'''Operations of the form <math>(L \,\setminus\, M) \ast (L \,\setminus\, R)</math>''':

<math display=block>\begin{alignat}{9}
(L \,\setminus M)  &\,\cup\,&&  (&&L \,\setminus R)  
            &&\;=\;&&  L \,\setminus\, (M \,\cap\, R)        \\[1.4ex]
(L \,\setminus M)  &\,\cap\,&&  (&&L \,\setminus R)  
            &&\;=\;&&  L \,\setminus\, (M \,\cup\, R)        \\[1.4ex]
(L \,\setminus M)  &\,\setminus\,&&  (&&L \,\setminus R)  
            &&\;=\;&&  (L \,\cap\, R) \,\setminus\, M        \\[1.4ex]
(L \,\setminus M)  &\,\triangle\,&&  (&&L \,\setminus R)  
            &&\;=\;&&  L \,\cap\, (M \,\triangle\, R)        \\[1.4ex]
&\,&&\,&&\, &&\;=\;&& (L \cap M)   \,\triangle\, (L \cap R)  \\[1.4ex]
\end{alignat}</math>

===Other simplifications===

'''Other properties''':

<math display=block>L \cap M = R \;\text{ and }\; L \cap R = M \qquad \text{ if and only if } \qquad M = R \subseteq L.</math> 

<ul>
<li>If <math>L \subseteq M</math> then <math>L \setminus R = L \cap (M \setminus R).</math>{{sfn|Császár|1978|pp=15-26}}</li>
<li><math>L \times (M \,\setminus R) = (L \times M) \,\setminus (L \times R)</math></li>
<li>If <math>L \subseteq R</math> then <math>M \setminus R \subseteq M \setminus L.</math></li>
<li><math>L \cap M \cap R = \varnothing</math> if and only if for any <math>x \in L \cup M \cup R,</math> <math>x</math> belongs to {{em|at most two}} of the sets <math>L, M, \text{ and } R.</math></li>
</ul>

==Cartesian products ⨯ of finitely many sets==

===Binary ⋂ of finite ⨯===

<math display=block>(L \times R) \cap \left(L_2 \times R_2\right) ~=~ \left(L \cap L_2\right) \times \left(R \cap R_2\right)</math>
<math display=block>(L \times M \times R) \cap \left(L_2 \times M_2 \times R_2\right) ~=~ \left(L \cap L_2\right) \times \left(M \cap M_2\right) \times \left(R \cap R_2\right)</math>

===Binary ⋃ of finite ⨯===

<math display=block>\begin{alignat}{9}
\left(L \times R\right) ~\cup~ \left(L_2 \times R_2\right) 
~&=~ \left[\left(L \setminus L_2\right) \times R\right] ~\cup~ \left[\left(L_2 \setminus L\right) \times R_2\right] ~\cup~ \left[\left(L \cap L_2\right) \times \left(R \cup R_2\right)\right] \\[0.5ex]
~&=~ \left[L \times \left(R \setminus R_2\right)\right] ~\cup~ \left[L_2 \times \left(R_2 \setminus R\right)\right] ~\cup~ \left[\left(L \cup L_2\right) \times \left(R \cap R_2\right)\right] \\
\end{alignat}</math>

===Difference \ of finite ⨯===

<math display=block>\begin{alignat}{9}
\left(L \times R\right) ~\setminus~ \left(L_2 \times R_2\right) 
~&=~ \left[\left(L \,\setminus\, L_2\right) \times R\right] ~\cup~ \left[L \times \left(R \,\setminus\, R_2\right)\right] \\
\end{alignat}</math>
and 
<math display=block>(L \times M \times R) ~\setminus~ \left(L_2 \times M_2 \times R_2\right) 
~=~ \left[\left(L \,\setminus\, L_2\right) \times M \times R\right] ~\cup~ \left[L \times \left(M \,\setminus\, M_2\right) \times R\right] ~\cup~ \left[L \times M \times \left(R \,\setminus\, R_2\right)\right]</math>

===Finite ⨯ of differences \===

<math display=block>\left(L \,\setminus\, L_2\right) \times \left(R \,\setminus\, R_2\right) ~=~ \left(L \times R\right) \,\setminus\, \left[\left(L_2 \times R\right) \cup \left(L \times R_2\right)\right]</math>

<math display=block>\left(L \,\setminus\, L_2\right) \times \left(M \,\setminus\, M_2\right) \times \left(R \,\setminus\, R_2\right) ~=~ \left(L \times M \times R\right) \,\setminus\, \left[\left(L_2 \times M \times R\right) \cup \left(L \times M_2 \times R\right) \cup \left(L \times M \times R_2\right)\right]</math>

===Symmetric difference ∆ and finite ⨯===

<math display=block>L \times \left(R \,\triangle\, R_2\right) ~=~ \left[L \times \left(R \,\setminus\, R_2\right)\right] \,\cup\, \left[L \times \left(R_2 \,\setminus\, R\right)\right]</math>
<math display=block>\left(L \,\triangle\, L_2\right) \times R ~=~ \left[\left(L \,\setminus\, L_2\right) \times R\right] \,\cup\, \left[\left(L_2 \,\setminus\, L\right) \times R\right]</math>

<math display=block>\begin{alignat}{4}
\left(L \,\triangle\, L_2\right) \times \left(R \,\triangle\, R_2\right) 
~&=~ && && \,\left[\left(L \cup L_2\right) \times \left(R \cup R_2\right)\right] \;\setminus\; \left[\left(\left(L \cap L_2\right) \times R\right) \;\cup\; \left(L \times \left(R \cap R_2\right)\right)\right] \\[0.7ex]
&=~ & &&& \,\left[\left(L \,\setminus\, L_2\right) \times \left(R_2 \,\setminus\, R\right)\right] \,\cup\, \left[\left(L_2 \,\setminus\, L\right) \times \left(R_2 \,\setminus\, R\right)\right] \,\cup\, \left[\left(L \,\setminus\, L_2\right) \times \left(R \,\setminus\, R_2\right)\right] \,\cup\, \left[\left(L_2 \,\setminus\, L\right) \cup \left(R \,\setminus\, R_2\right)\right] \\
\end{alignat}</math>

<math display=block>\begin{alignat}{4}
\left(L \,\triangle\, L_2\right) \times \left(M \,\triangle\, M_2\right) \times \left(R \,\triangle\, R_2\right) 
~&=~ \left[\left(L \cup L_2\right) \times \left(M \cup M_2\right) \times \left(R \cup R_2\right)\right] \;\setminus\; \left[\left(\left(L \cap L_2\right) \times M \times R\right) \;\cup\; \left(L \times \left(M \cap M_2\right) \times R\right) \;\cup\; \left(L \times M \times \left(R \cap R_2\right)\right)\right] \\
\end{alignat}</math>

In general, <math>\left(L \,\triangle\, L_2\right) \times \left(R \,\triangle\, R_2\right)</math> need not be a subset nor a superset of <math>\left(L \times R\right) \,\triangle\, \left(L_2 \times R_2\right).</math>

<math display=block>\begin{alignat}{4}
\left(L \times R\right) \,\triangle\, \left(L_2 \times R_2\right)
~&=~ && \left(L \times R\right) \cup \left(L_2 \times R_2\right) \;\setminus\; \left[\left(L \cap L_2\right) \times \left(R \cap R_2\right)\right] \\[0.7ex]
\end{alignat}</math>

<math display=block>\begin{alignat}{4}
\left(L \times M \times R\right) \,\triangle\, \left(L_2 \times M_2 \times R_2\right)
~&=~ && \left(L \times M \times R\right) \cup \left(L_2 \times M_2 \times R_2\right) \;\setminus\; \left[\left(L \cap L_2\right) \times \left(M \cap M_2\right) \times \left(R \cap R_2\right)\right] \\[0.7ex]
\end{alignat}</math>

==Arbitrary families of sets==

Let <math>\left(L_i\right)_{i \in I},</math> <math>\left(R_j\right)_{j \in J},</math> and <math>\left(S_{i,j}\right)_{(i, j) \in I \times J}</math> be indexed [[Family of sets|families of sets]]. Whenever the assumption is needed, then all indexing sets, such as <math>I</math> and <math>J,</math> are assumed to be non-empty. 

===Definitions===

A {{em|[[Family of sets|family of sets]]}} or (more briefly) a {{em|family}} refers to a set whose elements are sets. 

An {{em|[[Indexed family|indexed family]] of sets}} is a function from some set, called its {{em|indexing set}}, into some family of sets. 
An indexed family of sets will be denoted by <math>\left(L_i\right)_{i \in I},</math> where this notation assigns the symbol <math>I</math> for the indexing set and for every index <math>i \in I,</math> assigns the symbol <math>L_i</math> to the value of the function at <math>i.</math> 
The function itself may then be denoted by the symbol <math>L_\bull,</math> which is obtained from the notation <math>\left(L_i\right)_{i \in I}</math> by replacing the index <math>i</math> with a bullet symbol <math>\bullet\,;</math> explicitly, <math>L_\bull</math> is the function:
<math display=block>\begin{alignat}{4}
L_\bull :\;&& I &&\;\to    \;& \left\{L_i : i \in I\right\} \\[0.3ex]
           && i &&\;\mapsto\;& L_i \\
\end{alignat}</math>
which may be summarized by writing <math>L_\bull = \left(L_i\right)_{i \in I}.</math> 

Any given indexed family of sets <math>L_\bull = \left(L_i\right)_{i \in I}</math> (which is a [[Function (mathematics)|function]]) can be canonically associated with its image/range <math>\operatorname{Im} L_\bull ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{L_i : i \in I\right\}</math> (which is a family of sets). 
Conversely, any given family of sets <math>\mathcal{B}</math> may be associated with the <math>\mathcal{B}</math>-indexed family of sets <math>(B)_{B \in \mathcal{B}},</math> which is technically the identity map <math>\mathcal{B} \to \mathcal{B}.</math> 
However, this is {{em|not}} a bijective correspondence because an indexed family of sets <math>L_\bull = \left(L_i\right)_{i \in I}</math> is {{em|not}} required to be injective (that is, there may exist distinct indices <math>i \neq j</math> such as <math>L_i = L_j</math>), which in particular means that it is possible for distinct indexed families of sets (which are functions) to be associated with the same family of sets (by having the same image/range). 

'''Arbitrary unions defined'''{{sfn|Monk|1969|pp=24-54}}
{{NumBlk||<math display=block>\bigcup_{i \in I} L_i ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{x ~:~ \text{ there exists } i \in I \text{ such that } x \in L_i\}</math>|{{EquationRef|Def. 1}}}}

If <math>I = \varnothing</math> then <math>\bigcup_{i \in \varnothing} L_i = \{x ~:~ \text{ there exists } i \in \varnothing \text{ such that } x \in L_i\} = \varnothing,</math> which is somethings called the {{em|nullary union convention}} (despite being called a convention, this equality follows from the definition). 

If <math>\mathcal{B}</math> is a family of sets then <math>\cup \mathcal{B}</math> denotes the set: 
<math display=block>\bigcup \mathcal{B} ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \bigcup_{B \in \mathcal{B}} B ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{x ~:~ \text{ there exists } B \in \mathcal{B} \text{ such that } x \in B\}.</math>

'''Arbitrary intersections defined'''

If <math>I \neq \varnothing</math> then{{sfn|Monk|1969|pp=24-54}}
{{NumBlk||<math display=block>\bigcap_{i \in I} L_i ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{x ~:~ x \in L_i \text{ for every } i \in I\} ~=~ \{x ~:~  \text{ for all } i, \text{ if } i \in I \text{ then } x \in L_i\}.</math>|{{EquationRef|Def. 2}}}}

If <math>\mathcal{B} \neq \varnothing</math> is a {{em|non-empty}} family of sets then <math>\cap \mathcal{B}</math> denotes the set: 
<math display=block>\bigcap \mathcal{B} ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \bigcap_{B \in B} B ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{x ~:~ x \in B \text{ for every } B \in \mathcal{B}\} ~=~ \{x ~:~  \text{ for all } B, \text{ if } B \in \mathcal{B} \text{ then } x \in B\}.</math>

'''Nullary intersections'''

If <math>I = \varnothing</math> then
<math display=block>\bigcap_{i \in \varnothing} L_i = \{x ~:~  \text{ for all } i, \text{ if } i \in \varnothing \text{ then } x \in L_i\}</math>
where every possible thing <math>x</math> in the universe [[Vacuous truth|vacuously]] satisfied the condition: "if <math>i \in \varnothing</math> then <math>x \in L_i</math>". Consequently, <math>{\textstyle\bigcap\limits_{i \in \varnothing}} L_i = \{x : \text{ true }\}</math> consists of {{em|everything}} in the universe. 

So if <math>I = \varnothing</math> and:
#if you are working in a [[Model theory|model]] in which there exists some universe '''{{em|set}}''' <math>X</math> then <math>{\textstyle\bigcap\limits_{i \in \varnothing}} L_i = \{x ~:~ x \in L_i \text{ for every } i \in \varnothing\} ~=~ X.</math> 
#otherwise, if you are working in a [[Model theory|model]] in which "the class of all things <math>x</math>" is not a set (by far the most common situation) then <math>{\textstyle\bigcap\limits_{i \in \varnothing}} L_i</math> is {{em|undefined}} because <math>{\textstyle\bigcap\limits_{i \in \varnothing}} L_i</math> consists of {{em|everything}}, which makes <math>{\textstyle\bigcap\limits_{i \in \varnothing}} L_i</math> a proper class and {{em|not}} a set. 

:'''Assumption''': Henceforth, whenever a formula requires some indexing set to be non-empty in order for an arbitrary intersection to be well-defined, then this will automatically be assumed without mention. 

A consequence of this is the following assumption/definition:

:A '''{{em|finite intersection}} of sets''' or an '''{{em|intersection of finitely many sets}}''' refers to the intersection of a finite collection of {{em|one or more}} sets. 

Some authors adopt the so called '''{{em|nullary intersection convention}}''', which is the convention that an empty intersection of sets is equal to some canonical set. In particular, if all sets are subsets of some set <math>X</math> then some author may declare that the empty intersection of these sets be equal to <math>X.</math> However, the nullary intersection convention is not as commonly accepted as the nullary union convention and this article will not adopt it (this is due to the fact that unlike the empty union, the value of the empty intersection depends on <math>X</math> so if there are multiple sets under consideration, which is commonly the case, then the value of the empty intersection risks becoming ambiguous).

'''Multiple index sets'''
<math display=block>\bigcup_{\stackrel{i \in I,}{j \in J}} S_{i,j} ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \bigcup_{(i, j) \in I \times J} S_{i,j}</math>
<math display=block>\bigcap_{\stackrel{i \in I,}{j \in J}} S_{i,j} ~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \bigcap_{(i, j) \in I \times J} S_{i,j}</math>

===Distributing unions and intersections===

====Binary ⋂ of arbitrary ⋃'s{{anchor|Distributing a binary intersection of arbitrary unions}}====

{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \cap R ~=~ \bigcup_{i \in I} \left(L_i \cap R\right)</math>|{{EquationRef|Eq. 3a}}}}
and{{sfn|Császár|1978|pp=15-26}}
{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \cap \left(\bigcup_{j \in J} R_j\right) ~=~ \bigcup_{\stackrel{i \in I,}{j \in J}} \left(L_i \cap R_j\right)</math>|{{EquationRef|Eq. 3b}}}}

* If all <math>\left(L_i\right)_{i \in I}</math> are pairwise disjoint and all <math>\left(R_j\right)_{j \in J}</math> are also pairwise disjoint, then so are all <math>\left(L_i \cap R_j\right)_{(i, j) \in I \times J}</math> (that is, if <math>(i, j) \neq \left(i_2, j_2\right)</math> then <math>\left(L_i \cap R_j\right) \cap \left(L_{i_2} \cap R_{j_2}\right) = \varnothing</math>).

{{anchor|Ranging over pairs of indices}}
* '''Importantly''', if <math>I = J</math> then in general, <math display=block>~\left(\bigcup_{i \in I} L_i\right) \cap \left(\bigcup_{i \in I} R_i\right) ~~\color{Red}{\neq}\color{Black}{}~~ \bigcup_{i \in I} \left(L_i \cap R_i\right)~</math> (an example of this is given below).  The single union on the right hand side {{em|must}} be over all pairs <math>(i, j) \in I \times I:</math> <math display=block>~\left(\bigcup_{i \in I} L_i\right) \cap \left(\bigcup_{i \in I} R_i\right) ~~=~~ \bigcup_{\stackrel{i \in I,}{j \in I}} \left(L_i \cap R_j\right).~</math> The same is usually true for other similar non-trivial set equalities and relations that depend on two (potentially unrelated) indexing sets <math>I</math> and <math>J</math> (such as {{EquationNote|Eq. 4b}} or {{EquationNote|Eq. 7g}}{{sfn|Császár|1978|pp=15-26}}). Two exceptions are {{EquationNote|Eq. 2c}} (unions of unions) and {{EquationNote|Eq. 2d}} (intersections of intersections), but both of these are among the most trivial of set equalities (although even for these equalities there is still something that must be proven<ref group=note name="DeducingIntersectionOfIntersections"/>).
* '''{{visible anchor|CounterExampleToEqualityInterOfUnionsWhenIequalsJ|CounterExampleToEqualityUnionOfIntersWhenIequalsJ|text=Example where equality fails}}''': Let <math>X \neq \varnothing</math> and let <math>I = \{1, 2\}.</math> Let <math>L_1 \colon= R_2 \colon= X</math> and let <math>L_2 \colon= R_1 \colon= \varnothing.</math> Then <math display=block>X = X \cap X = \left(L_1 \cup L_2\right) \cap \left(R_2 \cup R_2\right) = \left(\bigcup_{i \in I} L_i\right) \cap \left(\bigcup_{i \in I} R_i\right) ~\neq~ \bigcup_{i \in I} \left(L_i \cap R_i\right) = \left(L_1 \cap R_1\right) \cup \left(L_2 \cap R_2\right) = \varnothing \cup \varnothing = \varnothing.</math> Furthermore, <math display=block>\varnothing = \varnothing \cup \varnothing = \left(L_1 \cap L_2\right) \cup \left(R_2 \cap R_2\right) = \left(\bigcap_{i \in I} L_i\right) \cup \left(\bigcap_{i \in I} R_i\right) ~\neq~ \bigcap_{i \in I} \left(L_i \cup R_i\right) = \left(L_1 \cup R_1\right) \cap \left(L_2 \cup R_2\right) = X \cap X = X.</math>

====Binary ⋃ of arbitrary ⋂'s{{anchor|Distributing a binary union of arbitrary intersections}}====

{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \cup R ~=~ \bigcap_{i \in I} \left(L_i \cup R\right)</math>|{{EquationRef|Eq. 4a}}}}
and{{sfn|Császár|1978|pp=15-26}}
{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \cup \left(\bigcap_{j \in J} R_j\right) ~=~ \bigcap_{\stackrel{i \in I,}{j \in J}} \left(L_i \cup R_j\right)</math>|{{EquationRef|Eq. 4b}}}}

* '''Importantly''', if <math>I = J</math> then in general, <math display=block>~\left(\bigcap_{i \in I} L_i\right) \cup \left(\bigcap_{i \in I} R_i\right) ~~\color{Red}{\neq}\color{Black}{}~~ \bigcap_{i \in I} \left(L_i \cup R_i\right)~</math> (an example of this is given above).  The single intersection on the right hand side {{em|must}} be over all pairs <math>(i, j) \in I \times I:</math> <math display=block>~\left(\bigcap_{i \in I} L_i\right) \cup \left(\bigcap_{i \in I} R_i\right) ~~=~~ \bigcap_{\stackrel{i \in I,}{j \in I}} \left(L_i \cup R_j\right).~</math>

====Arbitrary ⋂'s and arbitrary ⋃'s{{anchor|Arbitrary intersections and arbitrary unions|Distributing arbitrary intersections and arbitrary unions}}====

=====Incorrectly distributing by swapping ⋂ and ⋃{{anchor|Incorrectly distributing arbitrary intersections and arbitrary unions}}=====
'''Naively swapping <math>\;{\textstyle\bigcup\limits_{i \in I}}\;</math> and <math>\;{\textstyle\bigcap\limits_{j \in J}}\;</math> may produce a different set'''{{anchor|Arbitrary unions/intersections and switching symbols}}

The following inclusion always holds:
{{NumBlk||<math display=block>\bigcup_{i \in I} \left(\bigcap_{j \in J} S_{i,j}\right) ~~\color{Red}{\subseteq}\color{Black}{}~~ \bigcap_{j \in J} \left(\bigcup_{i \in I} S_{i,j}\right)</math>|{{EquationRef|In. 1|2=Inclusion 1 ∪∩ is a subset of ∩∪}}|LnSty=1px dashed black}}

In general, equality need not hold and moreover, the right hand side depends on how for each fixed <math>i \in I,</math> the sets <math>\left(S_{i,j}\right)_{j \in J}</math> are labelled; and analogously, the left hand side depends on how for each fixed <math>j \in J,</math> the sets <math>\left(S_{i,j}\right)_{i \in I}</math> are labelled. An example demonstrating this is now given.

<ul>
<li>'''Example of dependence on labeling and failure of equality''': To see why equality need not hold when <math>\cup</math> and <math>\cap</math> are swapped, let <math>I \colon= J \colon= \{1, 2\},</math> and let <math>S_{11} = \{1, 2\},~S_{12} = \{1, 3\},~S_{21} = \{3, 4\},</math> and <math>S_{22} = \{2, 4\}.</math> Then <math display=block>\{1, 4\} = \{1\} \cup \{4\} = \left(S_{11} \cap S_{12}\right) \cup \left(S_{21} \cap S_{22}\right) = \bigcup_{i \in I} \left(\bigcap_{j \in J} S_{i,j}\right) ~\neq~ \bigcap_{j \in J} \left(\bigcup_{i \in I} S_{i,j}\right) = \left(S_{11} \cup S_{21}\right) \cap \left(S_{12} \cup S_{22}\right) = \{1, 2, 3, 4\}.</math> 
If <math>S_{11}</math> and <math>S_{21}</math> are swapped while <math>S_{12}</math> and <math>S_{22}</math> are unchanged, which gives rise to the sets <math>\hat{S}_{11} \colon= \{3, 4\},~\hat{S}_{12} \colon= \{1, 3\},~\hat{S}_{21} \colon= \{1, 2\},</math> and <math>\hat{S}_{22} \colon= \{2, 4\},</math> then 
<math display=block>\{2, 3\} = \{3\} \cup \{2\} = \left(\hat{S}_{11} \cap \hat{S}_{12}\right) \cup \left(\hat{S}_{21} \cap \hat{S}_{22}\right) = \bigcup_{i \in I} \left(\bigcap_{j \in J} \hat{S}_{i,j}\right) ~\neq~ \bigcap_{j \in J} \left(\bigcup_{i \in I} \hat{S}_{i,j}\right) = \left(\hat{S}_{11} \cup \hat{S}_{21}\right) \cap \left(\hat{S}_{12} \cup \hat{S}_{22}\right) = \{1, 2, 3, 4\}.</math> 
In particular, the left hand side is no longer <math>\{1, 4\},</math> which shows that the left hand side <math>{\textstyle\bigcup\limits_{i \in I}} \; {\textstyle\bigcap\limits_{j \in J}} S_{i,j}</math> depends on how the sets are labelled. 
If instead <math>S_{11}</math> and <math>S_{12}</math> are swapped while <math>S_{21}</math> and <math>S_{22}</math> are unchanged, which gives rise to the sets <math>\overline{S}_{11} \colon= \{1, 3\},~ \overline{S}_{12} \colon= \{1, 2\},~ \overline{S}_{21} \colon= \{3, 4\},</math> and <math>\overline{S}_{22} \colon= \{2, 4\},</math> then both the left hand side and right hand side are equal to <math>\{1, 4\},</math> which shows that the right hand side also depends on how the sets are labeled.</li>
</ul>

Equality in {{EquationNote|In. 1|Inclusion 1 ∪∩ is a subset of ∩∪}} can hold under certain circumstances, such as in {{EquationNote|Eq. 7e|7e}}, which is the special case where <math>\left(S_{i,j}\right)_{(i,j) \in I \times J}</math> is <math>\left(L_i \setminus R_j\right)_{(i,j) \in I \times J}</math> (that is, <math>S_{i,j} \colon= L_i \setminus R_j</math> with the same indexing sets <math>I</math> and <math>J</math>), or such as in {{EquationNote|Eq. 7f|7f}}, which is the special case where <math>\left(S_{i,j}\right)_{(i,j) \in I \times J}</math> is <math>\left(L_i \setminus R_j\right)_{(j, i) \in J \times I}</math> (that is, <math>\hat{S}_{j,i} \colon= L_i \setminus R_j</math> with the indexing sets <math>I</math> and <math>J</math> swapped).
For a correct formula that extends the distributive laws, an approach other than just switching <math>\cup</math> and <math>\cap</math> is needed.

=====Correct distributive laws{{anchor|Correctly distributing arbitrary intersections and arbitrary unions|Correct distribution laws}}=====

Suppose that for each <math>i \in I,</math> <math>J_i</math> is a non-empty index set and for each <math>j \in J_i,</math> let <math>T_{i,j}</math> be any set (for example, to apply this law to <math>\left(S_{i,j}\right)_{(i, j) \in I \times J},</math> use <math>J_i \colon= J</math> for all <math>i \in I</math> and use <math>T_{i,j} \colon= S_{i,j}</math> for all <math>i \in I</math> and all <math>j \in J_i = J</math>). Let
<math display=block>{\textstyle \prod} J_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \prod_{i \in I} J_i</math> 
denote the [[Cartesian product]], which can be interpreted as the set of all functions <math>f ~:~ I ~\to~ {\textstyle\bigcup\limits_{i \in I}} J_i</math> such that <math>f(i) \in J_i</math> for every <math>i \in I.</math> Such a function may also be denoted using the tuple notation <math>\left(f_i\right)_{i \in I}</math> where <math>f_i ~\stackrel{\scriptscriptstyle\text{def}}{=}~ f(i)</math> for every <math>i \in I</math> and conversely, a tuple <math>\left(f_i\right)_{i \in I}</math> is just notation for the function with domain <math>I</math> whose value at <math>i \in I</math> is <math>f_i;</math> both notations can be used to denote the elements of <math>{\textstyle \prod} J_{\bull}.</math> 
Then 
{{NumBlk||<math display=block>\bigcap_{i \in I} \left[\;\bigcup_{j \in J_i} T_{i,j}\right] = \bigcup_{f \in \prod J_{\bull}} \left[\;\bigcap_{i \in I} T_{i,f(i)}\right]</math>|{{EquationRef|5|2=Eq. 5 ∩∪ to ∪∩}}|LnSty=1px dashed black}}
{{NumBlk||<math display=block>\bigcup_{i \in I} \left[\;\bigcap_{j \in J_i} T_{i,j}\right] = \bigcap_{f \in \prod J_{\bull}} \left[\;\bigcup_{i \in I} T_{i,f(i)}\right]</math>|{{EquationRef|6|2=Eq. 6 ∪∩ to ∩∪}}|LnSty=1px dashed black}}

where <math>{\textstyle \prod} J_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ {\textstyle\prod\limits_{i \in I}} J_i.</math> 

=====Applying the distributive laws{{anchor|Application of the distributive laws}}=====

'''{{em|Example application}}''': In the particular case where all <math>J_i</math> are equal (that is, <math>J_i = J_{i_2}</math> for all <math>i, i_2 \in I,</math> which is the case with the family <math>\left(S_{i,j}\right)_{(i, j) \in I \times J},</math> for example), then letting <math>J</math> denote this common set, the Cartesian product will be <math>{\textstyle \prod} J_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ {\textstyle\prod\limits_{i \in I}} J_i = {\textstyle\prod\limits_{i \in I}} J = J^I,</math> which is the [[Function space|set of all functions]] of the form <math>f ~:~ I ~\to~ J.</math> The above set equalities {{EquationNote|5|2=Eq. 5 ∩∪ to ∪∩}} and {{EquationNote|6|2=Eq. 6 ∪∩ to ∩∪}}, respectively become:{{sfn|Monk|1969|pp=24-54}}
<math display=block>\bigcap_{i \in I} \; \bigcup_{j \in J} S_{i,j} = \bigcup_{f \in J^I} \; \bigcap_{i \in I} S_{i,f(i)}</math>
<math display=block>\bigcup_{i \in I} \; \bigcap_{j \in J} S_{i,j} = \bigcap_{f \in J^I} \; \bigcup_{i \in I} S_{i,f(i)}</math>

which when combined with {{EquationNote|In. 1|2=Inclusion 1 ∪∩ is a subset of ∩∪}} implies: 
<math display=block>\bigcup_{i \in I} \; \bigcap_{j \in J} S_{i,j}
~=~ \bigcap_{f \in J^I} \; \bigcup_{i \in I} S_{i,f(i)}
~~\color{Red}{\subseteq}\color{Black}{}~~  \bigcup_{g \in I^J} \; \bigcap_{j \in J} S_{g(j),j}
~=~ \bigcap_{j \in J} \; \bigcup_{i \in I} S_{i,j}</math>
where 
* on the left hand side, the indices <math>f \text{ and } i</math> range over <math>f \in J^I \text{ and } i \in I</math> (so the subscripts of <math>S_{i,f(i)}</math> range over <math>i \in I \text{ and } f(i) \in f(I) \subseteq J</math>) 
* on the right hand side, the indices <math>g \text{ and } j</math> range over <math>g \in I^J \text{ and } j \in J</math> (so the subscripts of <math>S_{g(j),j}</math> range over <math>j \in J \text{ and } g(j) \in g(J) \subseteq I</math>). 


'''{{em|Example application}}''': To apply the general formula to the case of <math>\left(C_k\right)_{k \in K}</math> and <math>\left(D_{l}\right)_{l \in L},</math> use <math>I \colon= \{1, 2\},</math> <math>J_1 \colon= K,</math> <math>J_2 \colon= L,</math> and let <math>T_{1,k} \colon= C_k</math> for all <math>k \in J_1</math> and let <math>T_{2,l} \colon= D_l</math> for all <math>l \in J_2.</math> 
Every map <math>f \in {\textstyle \prod} J_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ {\textstyle\prod\limits_{i \in I}} J_i = J_1 \times J_2 = K \times L</math> can be [[Bijection|bijectively]] identified with the pair <math>\left(f(1), f(2)\right) \in K \times L</math> (the inverse sends <math>(k,l) \in K \times L</math> to the map <math>f_{(k,l)} \in {\textstyle \prod} J_{\bull}</math> defined by <math>1 \mapsto k</math> and <math>2 \mapsto l;</math> this is technically just a change of notation). Recall that {{EquationNote|5|2=Eq. 5 ∩∪ to ∪∩}} was  
<math display=block>~\bigcap_{i \in I} \; \bigcup_{j \in J_i} T_{i,j} = \bigcup_{f \in {\textstyle \prod} J_{\bull}} \; \bigcap_{i \in I} T_{i,f(i)}.~</math>  
Expanding and simplifying the left hand side gives
<math display=block>\bigcap_{i \in I} \; \bigcup_{j \in J_i} T_{i,j}
= \left(\bigcup_{j \in J_1} T_{1,j}\right) \cap \left(\;\bigcup_{j \in J_2} T_{2,j}\right) 
= \left(\bigcup_{k \in K} T_{1,k}\right) \cap \left(\;\bigcup_{l \in L} T_{2,l}\right) 
= \left(\bigcup_{k \in K} C_k\right) \cap \left(\;\bigcup_{l \in L} D_l\right)
</math>
and doing the same to the right hand side gives:
<math display=block>\bigcup_{f \in \prod J_{\bull}} \; \bigcap_{i \in I} T_{i,f(i)}
= \bigcup_{f \in \prod J_{\bull}} \left(T_{1,f(1)} \cap T_{2,f(2)}\right) 
= \bigcup_{f \in \prod J_{\bull}} \left(C_{f(1)} \cap D_{f(2)}\right) 
= \bigcup_{(k,l) \in K \times L} \left(C_k \cap D_l\right) 
= \bigcup_{\stackrel{k \in K,}{l \in L}} \left(C_k \cap D_l\right).
</math>

Thus the general identity {{EquationNote|5|2=Eq. 5 ∩∪ to ∪∩}} reduces down to the previously given set equality {{EquationNote|Eq. 3b}}: 
<math display=block>\left(\bigcup_{k \in K} C_k\right) \cap \;\bigcup_{l \in L} D_l = \bigcup_{\stackrel{k \in K,}{l \in L}} \left(C_k \cap  D_l\right).</math>

===Distributing subtraction over ⋃ and ⋂{{anchor|Distributing subtraction over unions and intersections}}===

{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \;\setminus\; R ~=~ \bigcup_{i \in I} \left(L_i \;\setminus\; R\right)</math>|{{EquationRef|Eq. 7a}}}}
{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \;\setminus\; R ~=~ \bigcap_{i \in I} \left(L_i \;\setminus\; R\right)</math>|{{EquationRef|Eq. 7b}}}}
The next identities are known as [[De Morgan's laws]].{{sfn|Császár|1978|pp=15-26}}
{{NumBlk||<math display=block>L \;\setminus\; \bigcup_{j \in J} R_j ~=~ \bigcap_{j \in J} \left(L \;\setminus\; R_j\right) ~~\;~~ \text{ De Morgan's law }</math>|{{EquationRef|Eq. 7c}}}}
{{NumBlk||<math display=block>L \;\setminus\; \bigcap_{j \in J} R_j ~=~ \bigcup_{j \in J} \left(L \;\setminus\; R_j\right) ~~\;~~ \text{ De Morgan's law }</math>|{{EquationRef|Eq. 7d}}}}

The following four set equalities can be deduced from the equalities {{EquationNote|Eq. 7a|7a}} - {{EquationNote|Eq. 7d|7d}} above. 
{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \;\setminus\; \bigcup_{j \in J} R_j   ~=~   \bigcup_{i \in I} \left(\bigcap_{j \in J} \left(L_i \;\setminus\; R_j\right)\right)   ~=~    \bigcap_{j \in J} \left(\bigcup_{i \in I} \left(L_i \;\setminus\; R_j\right)\right)</math>|{{EquationRef|Eq. 7e}}}}
{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \;\setminus\; \bigcap_{j \in J} R_j   ~=~    \bigcup_{j \in J} \left(\bigcap_{i \in I} \left(L_i \;\setminus\; R_j\right)\right)   ~=~   \bigcap_{i \in I} \left(\bigcup_{j \in J} \left(L_i \;\setminus\; R_j\right)\right)</math>|{{EquationRef|Eq. 7f}}}}
{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \;\setminus\; \bigcap_{j \in J} R_j   ~=~   \bigcup_{\stackrel{i \in I,}{j \in J}} \left(L_i \;\setminus\; R_j\right)</math>|{{EquationRef|Eq. 7g}}}}
{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \;\setminus\; \bigcup_{j \in J} R_j   ~=~   \bigcap_{\stackrel{i \in I,}{j \in J}} \left(L_i \;\setminus\; R_j\right)</math>|{{EquationRef|Eq. 7h}}}}

In general, naively swapping <math>\;\cup\;</math> and <math>\;\cap\;</math> may produce a different set (see this note for more details). 
The equalities 
<math display=block>\bigcup_{i \in I} \; \bigcap_{j \in J} \left(L_i \setminus R_j\right) ~=~ \bigcap_{j \in J} \; \bigcup_{i \in I} \left(L_i \setminus R_j\right)
\quad \text{ and } \quad  
\bigcup_{j \in J} \; \bigcap_{i \in I} \left(L_i \setminus R_j\right) ~=~ \bigcap_{i \in I} \; \bigcup_{j \in J} \left(L_i \setminus R_j\right)</math> 
found in {{EquationNote|Eq. 7e}} and {{EquationNote|Eq. 7f}} are thus unusual in that they state exactly that swapping <math>\;\cup\;</math> and <math>\;\cap\;</math> will {{em|not}} change the resulting set.

===Commutativity and associativity of ⋃ and ⋂{{anchor|Commutativity and associativity of unions and intersections}}===

'''Commutativity''':{{sfn|Monk|1969|pp=24-54}}

<math display=block>\bigcup_{\stackrel{i \in I,}{j \in J}} S_{i,j} 
~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \bigcup_{(i, j) \in I \times J} S_{i,j} 
~=~ \bigcup_{i \in I} \left(\bigcup_{j \in J} S_{i,j}\right) 
~=~ \bigcup_{j \in J} \left(\bigcup_{i \in I} S_{i,j}\right)</math>

<math display=block>\bigcap_{\stackrel{i \in I,}{j \in J}} S_{i,j} 
~~\stackrel{\scriptscriptstyle\text{def}}{=}~ \bigcap_{(i, j) \in I \times J} S_{i,j} 
~=~ \bigcap_{i \in I} \left(\bigcap_{j \in J} S_{i,j}\right) 
~=~ \bigcap_{j \in J} \left(\bigcap_{i \in I} S_{i,j}\right)</math>

'''Unions of unions and intersections of intersections''':{{sfn|Monk|1969|pp=24-54}}

<math display=block>\left(\bigcup_{i \in I} L_i\right) \cup R ~=~ \bigcup_{i \in I} \left(L_i \cup R\right)</math>
<math display=block>\left(\bigcap_{i \in I} L_i\right) \cap R ~=~ \bigcap_{i \in I} \left(L_i \cap R\right)</math>
and{{sfn|Monk|1969|pp=24-54}}
{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \cup \left(\bigcup_{j \in J} R_j\right) ~=~ \bigcup_{\stackrel{i \in I,}{j \in J}} \left(L_i \cup R_j\right)</math>|{{EquationRef|Eq. 2a}}}}
{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \cap \left(\bigcap_{j \in J} R_j\right) ~=~ \bigcap_{\stackrel{i \in I,}{j \in J}} \left(L_i \cap R_j\right)</math>|{{EquationRef|Eq. 2b}}}}

and if <math>I = J</math> then also:<ref group=note name="DeducingIntersectionOfIntersections">To deduce {{EquationNote|Eq. 2c}} from {{EquationNote|Eq. 2a}}, it must still be shown that <math>{\textstyle\bigcup\limits_{\stackrel{i \in I,}{j \in I}}} \left(L_i \cup R_j\right) ~=~ {\textstyle\bigcup\limits_{i \in I}} \left(L_i \cup R_i\right)</math> so {{EquationNote|Eq. 2c}} is not a completely immediate consequence of {{EquationNote|Eq. 2a}}. (Compare this to the commentary about {{EquationNote|Eq. 3b}}).</ref>{{sfn|Monk|1969|pp=24-54}}
{{NumBlk||<math display=block>\left(\bigcup_{i \in I} L_i\right) \cup \left(\bigcup_{i \in I} R_i\right) ~=~ \bigcup_{i \in I} \left(L_i \cup R_i\right)</math>|{{EquationRef|Eq. 2c}}}}
{{NumBlk||<math display=block>\left(\bigcap_{i \in I} L_i\right) \cap \left(\bigcap_{i \in I} R_i\right) ~=~ \bigcap_{i \in I} \left(L_i \cap R_i\right)</math>|{{EquationRef|Eq. 2d}}}}

===Cartesian products Π of arbitrarily many sets===

====Intersections ⋂ of Π====

If <math>\left(S_{i,j}\right)_{(i,j) \in I \times J}</math> is a family of sets then 
{{NumBlk||<math display=block>\bigcap_{j \in J} \; \prod_{i \in I} S_{i,j} ~~=~~ \prod_{i \in I} \; \bigcap_{j \in J} S_{i,j}</math>|{{EquationRef|Eq. 8}}}}
* Moreover, a tuple <math>\left(x_i\right)_{i \in I}</math> belongs to the set in {{EquationNote|Eq. 8|Eq. 8}} above if and only if <math>x_i \in S_{i,j}</math> for all <math>i \in I</math> and all <math>j \in J.</math>

In particular, if <math>\left(L_i\right)_{i \in I}</math> and <math>\left(R_i\right)_{i \in I}</math> are two families indexed by the same set then 
<math display=block>\left(\prod_{i \in I} L_i\right) \cap \prod_{i \in I} R_i ~=~ \prod_{i \in I} \left(L_i \cap R_i\right)</math>
So for instance, 
<math display=block>(L \times R) \cap \left(L_2 \times R_2\right) ~=~ \left(L \cap L_2\right) \times \left(R \cap R_2\right)</math>
<math display=block>(L \times R) \cap \left(L_2 \times R_2\right) \cap \left(L_3 \times R_3\right) ~=~ \left(L \cap L_2 \cap L_3\right) \times \left(R \cap R_2 \cap R_3\right)</math> and 
<math display=block>(L \times M \times R) \cap \left(L_2 \times M_2 \times R_2\right) ~=~ \left(L \cap L_2\right) \times \left(M \cap M_2\right) \times \left(R \cap R_2\right)</math>

'''Intersections of products indexed by different sets'''

Let <math>\left(L_i\right)_{i \in I}</math> and <math>\left(R_j\right)_{j \in J}</math> be two families indexed by different sets. 

Technically, <math>I \neq J</math> implies <math>\left({\textstyle\prod\limits_{i \in I}} L_i\right) \cap {\textstyle\prod\limits_{j \in J}} R_j = \varnothing.</math> 
However, sometimes these products are somehow identified as the same set through some [[Bijection|bijection]] or one of these products is identified as a subset of the other via some injective map, in which case (by [[Abuse of notation|abuse of notation]]) this intersection may be equal to some other (possibly non-empty) set. 
* For example, if <math>I := \{1, 2\}</math> and <math>J := \{1, 2, 3\}</math> with all sets equal to <math>\R</math> then <math>{\textstyle\prod\limits_{i \in I}} L_i = {\textstyle\prod\limits_{i \in \{1, 2\}}} \R = \R^2</math> and <math>{\textstyle\prod\limits_{j \in J}} R_j = {\textstyle\prod\limits_{j \in \{1, 2, 3\}}} \R = \R^3</math> where <math>\R^2 \cap \R^3 = \varnothing</math> {{em|unless}}, for example, <math>{\textstyle\prod\limits_{i \in \{1, 2\}}} \R = \R^2</math> is identified as a subset of <math>{\textstyle\prod\limits_{j \in \{1, 2, 3\}}} \R = \R^3</math> through some injection, such as maybe <math>(x, y) \mapsto (x, y, 0)</math> for instance; however, in this particular case the product <math>{\textstyle\prod\limits_{i \in I = \{1, 2\}}} L_i</math> actually represents the <math>J</math>-indexed product <math>{\textstyle\prod\limits_{j \in J = \{1, 2, 3\}}} L_i</math> where <math>L_3 := \{0\}.</math>
* For another example, take <math>I := \{1, 2\}</math> and <math>J := \{1, 2, 3\}</math> with <math>L_1 := \R^2</math> and <math>L_2, R_1, R_2, \text{ and } R_3</math> all equal to <math>\R.</math> Then <math>{\textstyle\prod\limits_{i \in I}}L_i = \R^2 \times \R</math> and <math>{\textstyle\prod\limits_{j \in J}} R_j = \R \times \R \times \R,</math> which can both be identified as the same set via the bijection that sends <math>((x, y), z) \in \R^2 \times \R</math> to <math>(x, y, z) \in \R \times \R \times \R.</math> Under this identification, <math>\left({\textstyle\prod\limits_{i \in I}} L_i\right) \cap \, {\textstyle\prod\limits_{j \in J}} R_j ~=~ \R^3.</math>

====Unions ⋃ of Π====

For unions, only the following is guaranteed in general:
<math display=block>\bigcup_{j \in J} \; \prod_{i \in I} S_{i,j} ~~\color{Red}{\subseteq}\color{Black}{}~~ \prod_{i \in I} \; \bigcup_{j \in J} S_{i,j} \qquad \text{ and } \qquad \bigcup_{i \in I} \; \prod_{j \in J} S_{i,j} ~~\color{Red}{\subseteq}\color{Black}{}~~ \prod_{j \in J} \; \bigcup_{i \in I} S_{i,j}</math>
where <math>\left(S_{i,j}\right)_{(i,j) \in I \times J}</math> is a family of sets. 

However, 
<math display=block>\begin{alignat}{9}
\left(L \times R\right) ~\cup~ \left(L_2 \times R_2\right) 
~&=~ \left[\left(L \setminus L_2\right) \times R\right] ~\cup~ \left[\left(L_2 \setminus L\right) \times R_2\right] ~\cup~ \left[\left(L \cap L_2\right) \times \left(R \cup R_2\right)\right] \\[0.5ex]
~&=~ \left[L \times \left(R \setminus R_2\right)\right] ~\cup~ \left[L_2 \times \left(R_2 \setminus R\right)\right] ~\cup~ \left[\left(L \cup L_2\right) \times \left(R \cap R_2\right)\right] \\
\end{alignat}</math>

====Difference \ of Π====

If <math>\left(L_i\right)_{i \in I}</math> and <math>\left(R_i\right)_{i \in I}</math> are two families of sets then:
<math display=block>\begin{alignat}{9}
\left(\prod_{i \in I} L_i\right) ~\setminus~ \prod_{i \in I}R_i
~&=~ \;~ \bigcup_{j \in I} \; ~ \prod_{i \in I} \begin{cases}L_j \,\setminus\, R_j & \text{ if } i = j \\ L_i & \text{ if } i \neq j \\ \end{cases} \\[0.5ex]
~&=~ \;~ \bigcup_{j \in I} \; ~ \Big[\left(L_j \,\setminus\, R_j\right) ~\times~ \prod_{\stackrel{i \in I,}{j \neq i}} L_i\Big] \\[0.5ex]
~&=~ \bigcup_{\stackrel{j \in I,}{L_j \not\subseteq R_j}} \Big[\left(L_j \,\setminus\, R_j\right) ~\times~ \prod_{\stackrel{i \in I,}{j \neq i}} L_i\Big] \\[0.3ex]
\end{alignat}</math>
so for instance,
<math display=block>\begin{alignat}{9}
\left(L \times R\right) ~\setminus~ \left(L_2 \times R_2\right) 
~&=~ \left[\left(L \,\setminus\, L_2\right) \times R\right] ~\cup~ \left[L \times \left(R \,\setminus\, R_2\right)\right] \\
\end{alignat}</math>
and 
<math display=block>(L \times M \times R) ~\setminus~ \left(L_2 \times M_2 \times R_2\right) 
~=~ \left[\left(L \,\setminus\, L_2\right) \times M \times R\right] ~\cup~ \left[L \times \left(M \,\setminus\, M_2\right) \times R\right] ~\cup~ \left[L \times M \times \left(R \,\setminus\, R_2\right)\right]</math>

====Symmetric difference ∆ of Π====

<math display=block>\begin{alignat}{9}
\left(\prod_{i \in I} L_i\right) ~\triangle~ \left(\prod_{i \in I} R_i\right)
~&=~ \;~ \left(\prod_{i \in I} L_i\right) ~\cup~ \left(\prod_{i \in I} R_i\right) \;\setminus\; \prod_{i \in I} L_i \cap R_i \\[0.5ex]
\end{alignat}</math>

==Functions and sets==

Let <math>f : X \to Y</math> be any function. 

Let <math>L \text{ and } R</math> be completely arbitrary sets. Assume <math>A \subseteq X \text{ and } C \subseteq Y.</math> 

===Definitions===

Let <math>f : X \to Y</math> be any function, where we denote its {{em|domain}} <math>X</math> by <math>\operatorname{domain} f</math> and denote its {{em|codomain}} <math>Y</math> by <math>\operatorname{codomain} f.</math> 

Many of the identities below do not actually require that the sets be somehow related to <math>f</math>'s domain or codomain (that is, to <math>X</math> or <math>Y</math>) so when some kind of relationship is necessary then it will be clearly indicated. 
Because of this, in this article, if <math>L</math> is declared to be "{{em|any set}}," and it is not indicated that <math>L</math> must be somehow related to <math>X</math> or <math>Y</math> (say for instance, that it be a subset <math>X</math> or <math>Y</math>) then it is meant that <math>L</math> is truly arbitrary.<ref group=note>So for instance, it's even possible that <math>L \cap (X \cup Y) = \varnothing,</math> or that <math>L \cap X \neq \varnothing</math> {{em|and}} <math>L \cap Y \neq \varnothing</math> (which happens, for instance, if <math>X = Y</math>), etc.</ref> 
This generality is useful in situations where <math>f : X \to Y</math> is a map between two subsets <math>X \subseteq U</math> and <math>Y \subseteq V</math> of some larger sets <math>U</math> and <math>V,</math> and where the set <math>L</math> might not be entirely contained in <math>X = \operatorname{domain} f</math> and/or <math>Y = \operatorname{codomain} f</math> (e.g. if all that is known about <math>L</math> is that <math>L \subseteq U</math>); in such a situation it may be useful to know what can and cannot be said about <math>f(L)</math> and/or <math>f^{-1}(L)</math> without having to introduce a (potentially unnecessary) intersection such as: <math>f(L \cap X)</math> and/or <math>f^{-1}(L \cap Y).</math>  

'''Images and preimages of sets'''

If <math>L</math> is {{em|any}} set then the {{em|[[Image (mathematics)|{{em|image}}]] of <math>L</math> under <math>f</math>}} is defined to be the set: 
<math display=block>f(L) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{\,f(l) ~:~ l \in L \cap \operatorname{domain} f\,\}</math>
while the {{em|preimage of <math>L</math> under <math>f</math>}} is: 
<math display=block>f^{-1}(L) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{\,x \in \operatorname{domain} f ~:~ f(x) \in L\,\}</math>
where if <math>L = \{s\}</math> is a singleton set then the {{em|[[Fiber (mathematics)|filber]]}} or {{em|preimage of <math>s</math> under <math>f</math>}} is
<math display=block>f^{-1}(s) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ f^{-1}(\{s\}) ~=~ \{\,x \in \operatorname{domain} f ~:~ f(x) = s\,\}.</math>

Denote by <math>\operatorname{Im} f</math> or <math>\operatorname{image} f</math> the {{em|image}} or {{em|[[Range of a function|range]]}} of <math>f : X \to Y,</math> which is the set: 
<math display=block>\operatorname{Im} f ~\stackrel{\scriptscriptstyle\text{def}}{=}~ f(X) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ f(\operatorname{domain} f) ~=~ \{f(x) ~:~ x \in \operatorname{domain} f\}.</math>

'''Saturated sets'''

{{Main|Saturated set}}

A set <math>A</math> is said to be {{em|<math>f</math>-'''saturated'''}} or a '''{{em|[[Saturated set|{{visible anchor|saturated set}}]]}}''' if any of the following equivalent conditions are satisfied:{{sfn|Monk|1969|pp=24-54}}  
<ol>
<li>There exists a set <math>R</math> such that <math>A = f^{-1}(R).</math>
* Any such set <math>R</math> necessarily contains <math>f(A)</math> as a subset.
* Any set not entirely contained in the domain of <math>f</math> cannot be <math>f</math>-saturated.</li>
<li><math>A = f^{-1}(f(A)).</math></li>
<li><math>A \supseteq f^{-1}(f(A))</math> and <math>A \subseteq \operatorname{domain} f.</math>
* The inclusion <math>L \cap \operatorname{domain} f \subseteq f^{-1}(f(L))</math> always holds, where if <math>A \subseteq \operatorname{domain} f</math> then this becomes <math>A \subseteq f^{-1}(f(A)).</math></li>
<li><math>A \subseteq \operatorname{domain} f</math> and if <math>a \in A</math> and <math>x \in \operatorname{domain} f</math> satisfy <math>f(x) = f(a),</math> then <math>x \in A.</math>
<li>Whenever a fiber of <math>f</math> intersects <math>A,</math> then <math>A</math> contains the entire fiber. In other words, <math>A</math> contains every <math>f</math>-fiber that intersects it.</li>
* Explicitly: whenever <math>y \in \operatorname{Im} f</math> is such that <math>A \cap f^{-1}(y) \neq \varnothing,</math> then <math>f^{-1}(y) s\subseteq A.</math>
* In both this statement and the next, the set <math>\operatorname{Im} f</math> may be replaced with any superset of <math>\operatorname{Im} f</math> (such as <math>\operatorname{codomain} f</math>) and the resulting statement will still be equivalent to the rest.
<li>The intersection of <math>A</math> with a fiber of <math>f</math> is equal to the empty set or to the fiber itself. 
* Explicitly: for every <math>y \in \operatorname{Im} f,</math> the intersection <math>A \cap f^{-1}(y)</math> is equal to the [[Empty set|empty set]] <math>\varnothing</math> or to <math>f^{-1}(y)</math> (that is, <math>A \cap f^{-1}(y) = \varnothing</math> or <math>A \cap f^{-1}(y) = f^{-1}(y)</math>). 
</li>
</ol>
For a set <math>A</math> to be <math>f</math>-saturated, it is necessary that <math>A \subseteq \operatorname{domain} f.</math> 

'''Compositions and restrictions of functions'''

If <math>f</math> and <math>g</math> are maps then <math>g \circ f</math> denotes the {{em|[[Function composition|composition]]}} map 
<math display=block>g \circ f ~:~ \{\,x \in \operatorname{domain} f ~:~ f(x) \in \operatorname{domain} g\,\} ~\to~ \operatorname{codomain} g</math>
with domain and codomain
<math display=block>\begin{alignat}{4}
\operatorname{domain}   (g \circ f) &= \{\,x \in \operatorname{domain} f ~:~ f(x) \in \operatorname{domain}  g\,\} \\[0.4ex]
\operatorname{codomain} (g \circ f) &= \operatorname{codomain} g \\[0.7ex]
\end{alignat}</math> 
defined by
<math display=block>(g \circ f)(x) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ g(f(x)).</math>

The {{em|[[Restriction (mathematics)|restriction]] of <math>f : X \to Y</math> to <math>L,</math>}} denoted by <math>f\big\vert_L,</math> is the map 
<math display=block>f\big\vert_L ~:~ L \cap \operatorname{domain} f ~\to~ Y</math>
with <math>\operatorname{domain} f\big\vert_L ~=~ L \cap \operatorname{domain} f</math> defined by sending <math>x \in L \cap \operatorname{domain} f</math> to <math>f(x);</math> that is, 
<math display=block>f\big\vert_L(x) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ f (x).</math> 
Alternatively, <math>~f\big\vert_L ~=~ f \circ \operatorname{In}~</math> where <math>~\operatorname{In} ~:~ L \cap X \to X~</math> denotes the [[Inclusion map|inclusion map]], which is defined by <math>\operatorname{In}(s) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s.</math>

===(Pre)Images of arbitrary unions ⋃'s and intersections ⋂'s===

If <math>\left(L_i\right)_{i \in I}</math> is a family of arbitrary sets indexed by <math>I \neq \varnothing</math> then:{{sfn|Császár|1978|pp=102-120}}
<math display=block>\begin{alignat}{4}
f\left(\bigcap_{i \in I} L_i\right)       \;&~\;\color{Red}{\subseteq}\color{Black}{}~ \;\;\;\bigcap_{i \in I} f\left(L_i\right) \\
f\left(\bigcup_{i \in I} L_i\right)       \;&~=~ \;\bigcup_{i \in I} f\left(L_i\right) \\
f^{-1}\left(\bigcup_{i \in I} L_i\right)  \;&~=~ \;\bigcup_{i \in I} f^{-1}\left(L_i\right) \\
f^{-1}\left(\bigcap_{i \in I} L_i\right)  \;&~=~ \;\bigcap_{i \in I} f^{-1}\left(L_i\right) \\
\end{alignat}</math>

So of these four identities, it is {{em|<u>only</u>}} '''{{em|images of intersections}}''' that are not always preserved. Preimages preserve all basic set operations. Unions are preserved by both images and preimages. 

If all <math>L_i</math> are <math>f</math>-saturated then <math>\bigcap_{i \in I} L_i</math> be will be <math>f</math>-saturated and equality will hold in the first relation above; explicitly, this means: 
{{NumBlk||<math display=block>f\left(\bigcap_{i \in I} L_i\right)  ~=~  \bigcap_{i \in I} f\left(L_i\right) \qquad \textit{IF} \qquad X \cap L_i = f^{-1}\left(f\left(L_i\right)\right) \quad \text{ for all } \quad i \in I.</math>|{{EquationRef|Conditional Equality 10a}}}}

If <math>\left(A_i\right)_{i \in I}</math> is a family of arbitrary subsets of <math>X = \operatorname{domain} f,</math> which means that <math>A_i \subseteq X</math> for all <math>i,</math> then {{EquationNote|Conditional Equality 10a}} becomes:
{{NumBlk||<math display=block>f\left(\bigcap_{i \in I} A_i\right)  ~=~  \bigcap_{i \in I} f\left(A_i\right) \qquad \textit{IF} \qquad A_i = f^{-1}\left(f\left(A_i\right)\right) \quad \text{ for all } \quad i \in I.</math>|{{EquationRef|Conditional Equality 10b}}}}

===(Pre)Images of binary set operations===

Throughout, let <math>L</math> and <math>R</math> be any sets and let <math>f : X \to Y</math> be any function.

'''Summary'''

As the table below shows, set equality is {{em|not}} guaranteed {{em|only}} for {{em|images}} of: intersections, set subtractions, and symmetric differences. 

{| class="wikitable"
|-
! Image
! Preimage
! {{nowrap|Additional assumptions on sets}}

|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\,~~~~f(L \cup R) ~=~ f(L) \cup f(R)</math><ref name="kelley-1985">{{harvnb|Kelley|1985|p=[{{Google books|plainurl=y|id=-goleb9Ov3oC|page=85|text=The image of the union of a family of subsets of X is the union of the images, but, in general, the image of the intersection is not the intersection of the images}} 85]}}</ref>
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(L \cup R) ~=~ f^{-1}(L) \cup f^{-1}(R)</math>{{sfn|Monk|1969|pp=24-54}}
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math display=block>f(L \cap R) ~\subseteq~ f(L) \cap f(R)</math>
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(L \cap R) ~=~ f^{-1}(L) \cap f^{-1}(R)</math>{{sfn|Monk|1969|pp=24-54}}
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math display=block>f(L \setminus R) ~\supseteq~ f(L) \setminus f(R)</math>
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
f^{-1}(L) \setminus f^{-1}(R) 
&= f^{-1}&&( &&L                           &&\setminus  &&R) \\
&= f^{-1}&&( &&L                           &&\setminus [&&R \cap \operatorname{Im} f]) \\
&= f^{-1}&&([&&L \cap \operatorname{Im} f] &&\setminus  &&R) \\
&= f^{-1}&&([&&L \cap \operatorname{Im} f] &&\setminus [&&R \cap \operatorname{Im} f]) 
\end{alignat}</math>{{sfn|Császár|1978|pp=102-120}}{{sfn|Monk|1969|pp=24-54}}
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math display=block>f(X \setminus R) ~\supseteq~ f(X) \setminus f(R)</math>
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
X \setminus f^{-1}(R) 
&= f^{-1}(&&Y                   &&\setminus  &&R) \\
&= f^{-1}(&&Y                   &&\setminus [&&R \cap \operatorname{Im} f]) \\
&= f^{-1}(&&\operatorname{Im} f &&\setminus  &&R) \\
&= f^{-1}(&&\operatorname{Im} f &&\setminus [&&R \cap \operatorname{Im} f]) 
\end{alignat}</math><ref group=note>The conclusion <math>X \setminus f^{-1}(R) = f^{-1}(Y \setminus R)</math> can also be written as: <math>f^{-1}(R)^\complement ~=~ f^{-1}\left(R^\complement\right).</math></ref>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math display=block>f\left(L ~\triangle~ R\right) ~\supseteq~ f(L) ~\triangle~ f(R)</math>
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}\left(L ~\triangle~ R\right) ~=~ f^{-1}(L) ~\triangle~ f^{-1}(R)</math>
|None
|}

'''Preimages preserve set operations'''

Preimages of sets are well-behaved with respect to all basic set operations:

<math display=block>\begin{alignat}{4}
f^{-1}(L \cup R) ~&=~ f^{-1}(L) \cup f^{-1}(R) \\
f^{-1}(L \cap R) ~&=~ f^{-1}(L) \cap f^{-1}(R) \\
f^{-1}(L \setminus\, R) ~&=~ f^{-1}(L) \setminus\, f^{-1}(R) \\
f^{-1}(L \,\triangle\, R) ~&=~ f^{-1}(L) \,\triangle\, f^{-1}(R) \\
\end{alignat}</math>

In words, preimages [[Distributive property|distribute over]] unions, intersections, set subtraction, and symmetric difference. 

'''Images {{em|only}} preserve unions'''

Images of unions are well-behaved:

<math display=block>\begin{alignat}{4}
f(L \cup R) ~&=~ f(L) \cup f(R) \\
\end{alignat}</math>

but images of the other basic set operations are {{em|not}} since only the following are guaranteed in general:

<math display=block>\begin{alignat}{4}
f(L \cap R) ~&\subseteq~ f(L) \cap f(R) \\
f(L \setminus R) ~&\supseteq~ f(L) \setminus f(R) \\
f(L \triangle R) ~&\supseteq~ f(L) \,\triangle\, f(R) \\
\end{alignat}</math>

In words, images [[Distributive property|distribute over]] unions but not necessarily over intersections, set subtraction, or symmetric difference. What these latter three operations have in common is set subtraction: they either {{em|are}} set subtraction <math>L \setminus R</math> or else they can naturally {{em|be defined}} as the set subtraction of two sets: 
<math display=block>L \cap R = L \setminus (L \setminus R) \quad \text{ and } \quad L \triangle R = (L \cup R) \setminus (L \cap R).</math> 

If <math>L = X</math> then <math>f(X \setminus R) \supseteq f(X) \setminus f(R)</math> where as in the more general case, equality is not guaranteed. If <math>f</math> is surjective then <math>f(X \setminus R) ~\supseteq~ Y \setminus f(R),</math> which can be rewritten as: <math>f\left(R^\complement\right) ~\supseteq~ f(R)^\complement</math> if <math>R^\complement ~\stackrel{\scriptscriptstyle\text{def}}{=}~ X \setminus R</math> and <math>f(R)^\complement ~\stackrel{\scriptscriptstyle\text{def}}{=}~ Y \setminus f(R).</math>

====Counter-examples: images of operations not distributing====

[[File:Image preimage conterexample intersection.gif|thumb|right|upright=1.6|Picture showing <math>f</math> failing to [[Distributive property|distribute over set intersection]]:<br/>
{{center|<math>f\left(A_1 \cap A_2\right) \subsetneq f\left(A_1\right) \cap f\left(A_2\right).</math>}}
The map <math>f : \R \to \R</math> is defined by <math>x \mapsto x^2,</math> where <math>\R</math> denotes the [[Real number|real number]]s. The sets <math>A_1 = [-4, 2]</math> and <math>A_2 = [-2, 4]</math> are shown in {{color|blue|blue}} immediately below the <math>x</math>-axis while their intersection <math>A_3 = [-2, 2]</math> is shown in {{color|green|green}}.]]

If <math>f : \{1, 2\} \to Y</math> is constant, <math>L = \{1\},</math> and <math>R = \{2\}</math> then all four of the set containments
<math display=block>\begin{alignat}{4}
f(L \cap R) ~&\subsetneq~ f(L) \cap f(R) \\
f(L \setminus R) ~&\supsetneq~ f(L) \setminus f(R) \\
f(X \setminus R) ~&\supsetneq~ f(X) \setminus f(R) \\
f(L \triangle R) ~&\supsetneq~ f(L) \triangle f(R) \\
\end{alignat}</math>
are strict/proper (that is, the sets are not equal) since one side is the empty set while the other is non-empty. Thus equality is not guaranteed for even the simplest of functions. 
The example above is now generalized to show that these four set equalities can fail for any [[Constant function|constant function]] whose domain contains at least two (distinct) points. 

'''{{em|Example}}''':{{anchor|ConstantFunctionWithDisjointSubsetsExample}} Let <math>f : X \to Y</math> be any constant function with image <math>f(X) = \{y\}</math> and suppose that <math>L, R \subseteq X</math> are non-empty disjoint subsets; that is, <math>L \neq \varnothing, R \neq \varnothing,</math> and <math>L \cap R = \varnothing,</math> which implies that all of the sets <math>L ~\triangle~ R = L \cup R,</math> <math>\,L \setminus R = L,</math> and <math>X \setminus R \supseteq L \setminus R</math> are not empty and so consequently, their images under <math>f</math> are all equal to <math>\{y\}.</math> 

<ol>
<li>The containment <math>~f(L \setminus R) ~\supsetneq~ f(L) \setminus f(R)~</math> is strict: <math display=block>\{y\} ~=~ f(L \setminus R) ~\neq~ f(L) \setminus f(R) ~=~ \{y\} \setminus \{y\} ~=~ \varnothing</math>
In words: functions might not [[Distributive property|distribute over set subtraction <math>\,\setminus\,</math>]] 
</li>
<li>The containment <math>~f(X \setminus R) ~\supsetneq~ f(X) \setminus f(R)~</math> is strict: <math display=block>\{y\} ~=~ f(X \setminus R) ~\neq~ f(X) \setminus f(R) ~=~ \{y\} \setminus \{y\} ~=~ \varnothing.</math>
</li>
<li>The containment <math>~f(L ~\triangle~ R) ~\supsetneq~ f(L) ~\triangle~ f(R)~</math> is strict: <math display=block>\{y\} ~=~ f\left(L ~\triangle~ R\right) ~\neq~ f(L) ~\triangle~ f(R) ~=~ \{y\} \triangle \{y\} ~=~ \varnothing</math>
In words: functions might not [[Distributive property|distribute over symmetric difference <math>\,\triangle\,</math>]] (which can be defined as the set subtraction of two sets: <math>L \triangle R = (L \cup R) \setminus (L \cap R)</math>).
</li>
<li>The containment <math>~f(L \cap R) ~\subsetneq~ f(L) \cap f(R)~</math> is strict: <math display=block>\varnothing ~=~ f(\varnothing) ~=~ f(L \cap R) ~\neq~ f(L) \cap f(R) ~=~ \{y\} \cap \{y\} ~=~ \{y\}</math>
In words: functions might not [[Distributive property|distribute over set intersection <math>\,\cap\,</math>]] (which can be defined as the set subtraction of two sets: <math>L \cap R = L \setminus (L \setminus R)</math>).
</li>
</ol>

What the set operations in these four examples have in common is that they either {{em|are}} set subtraction <math>\setminus</math> (examples (1) and (2)) or else they can naturally {{em|be defined}} as the set subtraction of two sets (examples (3) and (4)).

'''Mnemonic''': In fact, for each of the above four set formulas for which equality is not guaranteed, the direction of the containment (that is, whether to use <math>\,\subseteq \text{ or } \supseteq\,</math>) can always be deduced by imagining the function <math>f</math> as being {{em|constant}} and the two sets (<math>L</math> and <math>R</math>) as being non-empty disjoint subsets of its domain. This is because {{em|every}} equality fails for such a function and sets: one side will be always be <math>\varnothing</math> and the other non-empty − from this fact, the correct choice of <math>\,\subseteq \text{ or } \supseteq\,</math> can be deduced by answering: "which side is empty?" For example, to decide if the <math>?</math> in 
<math display=block>f(L \triangle R) \setminus f(R) ~\;~?~\;~ f((L \triangle R) \setminus R)</math>
should be <math>\,\subseteq \text{ or } \supseteq,\,</math> pretend<ref group=note>Whether or not it is even feasible for the function <math>f</math> to be constant and the sets <math>L \triangle R</math> and <math>R</math> to be non-empty and disjoint is irrelevant for reaching the correct conclusion about whether to use <math>\,\subseteq \text{ or } \supseteq.\,</math></ref> 
that <math>f</math> is constant and that <math>L \triangle R</math> and <math>R</math> are non-empty disjoint subsets of <math>f</math>'s domain; then the {{em|left}} hand side would be empty (since <math>f(L \triangle R) \setminus f(R) = \{f\text{'s single value}\} \setminus \{f\text{'s single value}\} = \varnothing</math>), which indicates that <math>\,?\,</math> should be <math>\,\subseteq\,</math> (the resulting statement is always guaranteed to be true) because this is the choice that will make
<math display=block>\varnothing = \text{left hand side} ~\;~?~\;~ \text{right hand side}</math>
true. 
Alternatively, the correct direction of containment can also be deduced by consideration of any constant <math>f : \{1, 2\} \to Y</math> with <math>L = \{1\}</math> and <math>R = \{2\}.</math> 

Furthermore, this mnemonic can also be used to correctly deduce whether or not a set operation always distribute over images or preimages; for example, to determine whether or not <math>f(L \cap R)</math> always equals <math>f(L) \cap f(R),</math> or alternatively, whether or not <math>f^{-1}(L \cap R)</math> always equals <math>f^{-1}(L) \cap f^{-1}(R)</math> (although <math>\,\cap\,</math> was used here, it can replaced by <math>\,\cup, \,\setminus, \text{ or } \,\triangle</math>). The answer to such a question can, as before, be deduced by consideration of this constant function: the answer for the general case (that is, for arbitrary <math>f, L,</math> and <math>R</math>) is always the same as the answer for this choice of (constant) function and disjoint non-empty sets.

====Conditions guaranteeing that images distribute over set operations====

'''Characterizations of when equality holds for {{em|all}} sets''':

For any function <math>f : X \to Y,</math> the following statements are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f : X \to Y</math> is [[Injective function|injective]]. 
* This means: <math>f(x) \neq f(y)</math> for all distinct <math>x, y \in X.</math>
</li>
<li><math>f(L \cap R) = f(L) \,\cap\, f(R) \, \text{ for all } \, L, R \subseteq X.</math> (The equals sign <math>\,=\,</math> can be replaced with <math>\,\supseteq\,</math>).
</li>
<li><math>f(L \,\setminus R) = f(L) \,\setminus\, f(R) \; \text{ for all } \, L, R \subseteq X.</math> (The equals sign <math>\,=\,</math> can be replaced with <math>\,\subseteq\,</math>).
</li>
<li><math>f(X \setminus R) = f(X) \setminus\, f(R) \; \text{ for all } \, ~~~~~R \subseteq X.</math> (The equals sign <math>\,=\,</math> can be replaced with <math>\,\subseteq\,</math>).
</li>
<li><math>f(L \,\triangle\, R) = f(L) \,\triangle\, f(R) \; \text{ for all } \, L, R \subseteq X.</math> (The equals sign <math>\,=\,</math> can be replaced with <math>\,\subseteq\,</math>).
</li>
<li>Any one of the four statements (b) - (e) but with the words "for all" replaced with any one of the following:
<ol style="list-style-type: lower-latin;">
<li>"for all singleton subsets"
* In particular, the statement that results from (d) gives a characterization of injectivity that explicitly involves only one point (rather than two): <math>f</math> is injective if and only if <math>f(x) \not\in f(X \setminus \{x\}) \; \text{ for every } \, x \in X.</math></li>
<li>"for all [[Disjoint sets|disjoint]] singleton subsets"
* For statement (d), this is the same as: "for all singleton subsets" (because the definition of "pairwise disjoint" is satisfies vacuously by any family that consists of exactly 1 set).</li>
<li>"for all disjoint subsets"</li>
</ol>
</li>
</ol>

In particular, if a map is not known to be injective then barring additional information, there is no guarantee that any of the equalities in statements (b) - (e) hold.

An example above can be used to help prove this characterization. Indeed, comparison of that example with such a proof suggests that the example is representative of the fundamental reason why one of these four equalities in statements (b) - (e) might not hold (that is, representative of "what goes wrong" when a set equality does not hold).

=====Conditions for f(L⋂R) = f(L)⋂f(R){{anchor|Image of an intersection ⋂|Image of an intersection}}=====

<math display=block>f(L \cap R) ~\subseteq~ f(L) \cap f(R) \qquad\qquad \text{ always holds}</math>

''Characterizations of equality'': The following statements are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f(L \cap R) ~=~ f(L) \cap f(R)</math></li>
<li><math>f(L \cap R) ~\supseteq~ f(L) \cap f(R)</math></li>
<li><math>L \cap f^{-1}(f(R)) ~\subseteq~ f^{-1}(f(L \cap R))</math>
* The left hand side <math>L \cap f^{-1}(f(R))</math> is always equal to <math>L \cap f^{-1}(f(L) \cap f(R))</math> (because <math>L \cap f^{-1}(f(R)) ~\subseteq~ f^{-1}(f(L))</math> always holds).</li>
<li><math>R \cap f^{-1}(f(L)) ~\subseteq~ f^{-1}(f(L \cap R))</math></li>
<li><math>L \cap f^{-1}(f(R)) ~=~ f^{-1}(f(L \cap R)) \cap L</math></li>
<li><math>R \cap f^{-1}(f(L)) ~=~ f^{-1}(f(L \cap R)) \cap R</math></li>
<li>If <math>l \in L</math> satisfies <math>f(l) \in f(R)</math> then <math>f(l) \in f(L \cap R).</math></li>
<li>If <math>y \in f(L)</math> but <math>y \notin f(L \cap R)</math> then <math>y \notin f(R).</math></li>
<li><math>f(L) \, \setminus \, f(L \cap R) ~\subseteq~ f(L) \, \setminus \, f(R)</math></li>
<li><math>f(R) \, \setminus \, f(L \cap R) ~\subseteq~ f(R) \, \setminus \, f(L)</math></li>
<li><math>f(L \cup R) \setminus f(L \cap R) ~\subseteq~ f(L) \,\triangle\, f(R)</math></li>
<li>Any of the above three conditions (i) - (k) but with the subset symbol <math>\,\subseteq\,</math> replaced with an equals sign <math>\,=.\,</math></li>
</ol>

''Sufficient conditions for equality'': Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>f</math> is injective.<ref name="munkres-2000-p21">See {{harvnb|Munkres|2000|p=21}}</ref></li>
<li>The restriction <math>f\big\vert_{L \cup R}</math> is injective.</li>
<li><math>f^{-1}(f(R)) ~\subseteq~ R</math><ref group=note name="Depends only on R"/></li>
<li><math>f^{-1}(f(L)) ~\subseteq~ L</math>
<!--This is equivalent to: <li><math>R \cap \operatorname{domain} f ~=~ f^{-1}(f(R))</math><ref group=note name="Depends only on R"/> &nbsp;or<br/><math>L \cap \operatorname{domain} f ~=~ f^{-1}(f(L))</math></li>--></li>
<li><math>R</math> is <math>f</math>-saturated; that is, <math>f^{-1}(f(R)) = R</math><ref group=note name="Depends only on R"/></li>
<li><math>L</math> is <math>f</math>-saturated; that is, <math>f^{-1}(f(L)) = L</math></li>
<li><math>f(L) ~\subseteq~ f(L \cap R)</math></li>
<li><math>f(R) ~\subseteq~ f(L \cap R)</math></li>
<li><math>f(L \,\setminus\, R) ~\subseteq~ f(L) \setminus \, f(R)</math> or equivalently, <math>f(L \,\setminus\, R) ~=~ f(L) \setminus f(R)</math></li>
<li><math>f(R \,\setminus\, L) ~\subseteq~ f(R) \setminus \, f(L)</math> or equivalently, <math>f(R \,\setminus\, L) ~=~ f(R) \setminus f(L)</math></li>
<li><math>f\left(L ~\triangle~ R\right) \subseteq f(L) ~\triangle~ f(R)</math> or equivalently, <math>f\left(L ~\triangle~ R\right) = f(L) ~\triangle~ f(R)</math></li>
<li><math>R \cap \operatorname{domain} f \,\subseteq L</math></li>
<li><math>L \cap \operatorname{domain} f \,\subseteq R</math></li>
<li><math>R \subseteq L</math></li>
<li><math>L \subseteq R</math></li>
</ol>

In addition, the following always hold:
<math display=block>f\left(f^{-1}(L) \cap R\right) ~=~ L \cap f(R)</math>
<math display=block>f\left(f^{-1}(L) \cup R\right) ~=~ (L \cap \operatorname{Im} f) \cup f(R)</math>

=====Conditions for f(L\R) = f(L)\f(R){{anchor|Image of a set subtraction \|Image of a set subtraction}}=====

<math display=block>f(L \setminus R) ~\supseteq~ f(L) \setminus f(R) \qquad\qquad \text{ always holds}</math>

''Characterizations of equality'': The following statements are equivalent:<ref group=proof name=proofCharOfPullingLminusR />
<ol style="list-style-type: lower-latin;">
<li><math>f(L \setminus R) ~=~ f(L) \setminus f(R)</math></li>
<li><math>f(L \setminus R) ~\subseteq~ f(L) \setminus f(R)</math></li>
<li><math>L \cap f^{-1}(f(R)) ~\subseteq~ R</math></li>
<li><math>L \cap f^{-1}(f(R)) ~=~ L \cap R \cap \operatorname{domain} f</math></li>
<li>Whenever <math>y \in f(L) \cap f(R)</math> then <math>L \cap f^{-1}(y) \subseteq R.</math></li>
<li><math display=inline>f(L) \cap f(R) ~\subseteq~ \left\{y \in f(L) : L \cap f^{-1}(y) \subseteq R\right\}</math>
* The set on the right hand side is always equal to <math>\left\{y \in f(L \cap R) :  L \cap f^{-1}(y) \,\subseteq R\right\}.</math></li>
<li><math display=inline>f(L) \cap f(R) ~=~ \left\{y \in f(L) : L \cap f^{-1}(y) \subseteq R\right\}</math>
* This is the above condition (f) but with the subset symbol <math>\,\subseteq\,</math> replaced with an equals sign <math>\,=.\,</math></li>
</ol>

''Necessary conditions for equality'' (excluding characterizations):  If equality holds then the following are necessarily true:
<ol style="list-style-type: lower-latin;">
<li><math>f(L \cap R) = f(L) \cap f(R),</math> or equivalently <math>f(L \cap R) \supseteq f(L) \cap f(R).</math></li>
<li><math>L \cap f^{-1}(f(R)) ~=~ L \cap f^{-1}(f(L \cap R))</math> or equivalently, <math>L \cap f^{-1}(f(R)) ~\subseteq~ f^{-1}(f(L \cap R))</math></li>
<li><math>R \cap f^{-1}(f(L)) ~=~ R \cap f^{-1}(f(L \cap R))</math></li>
</ol>

''Sufficient conditions for equality'': Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>f</math> is injective.</li>
<li>The restriction <math>f\big\vert_{L \cup R}</math> is injective.</li>
<li><math>f^{-1}(f(R)) ~\subseteq~ R</math><ref group=note name="Depends only on R">Note that this condition depends entirely on <math>R</math> and {{em|not}} on <math>L.</math></ref> or equivalently, <math>R \cap \operatorname{domain} f ~=~ f^{-1}(f(R))</math></li>
<li><math>R</math> is <math>f</math>-saturated; that is, <math>R = f^{-1}(f(R)).</math><ref group=note name="Depends only on R"/></li>
<li><math>f\left(L ~\triangle~ R\right) \subseteq f(L) ~\triangle~ f(R)</math> or equivalently, <math>f\left(L ~\triangle~ R\right) = f(L) ~\triangle~ f(R)</math></li>
</ol>

=====Conditions for f(X\R) = f(X)\f(R){{anchor|Image of a set subtracted from the domain \|Image of a set subtracted from the domain}}=====

<math display=block>f(X \setminus R) ~\supseteq~ f(X) \setminus f(R) \qquad\qquad \text{ always holds, where } f : X \to Y</math>

''Characterizations of equality'': The following statements are equivalent:<ref group=proof name=proofCharOfPullingLminusR />
<ol style="list-style-type: lower-latin;">
<li><math>f(X \setminus R) ~=~ f(X) \setminus f(R)</math></li>
<li><math>f(X \setminus R) ~\subseteq~ f(X) \setminus f(R)</math></li>
<li><math>f^{-1}(f(R)) \,\subseteq\, R</math></li>
<li><math>f^{-1}(f(R)) \,=\, R \cap \operatorname{domain} f</math></li>
<li><math>R \cap \operatorname{domain} f</math> is <math>f</math>-saturated.</li>
<li>Whenever <math>y \in f(R)</math> then <math>f^{-1}(y) \subseteq R.</math></li>
<li><math display=inline>f(R) ~\subseteq~ \left\{ y \in f(R) : f^{-1}(y) \subseteq R \right\}</math></li>
<li><math display=inline>f(R) ~=~ \left\{ y \in f(R) : f^{-1}(y) \subseteq R \right\}</math></li>
</ol>
&nbsp;&nbsp;&nbsp;where if <math>R \subseteq \operatorname{domain} f</math> then this list can be extended to include:
<ol style="list-style-type: lower-latin;" start=9>
<li><math>R</math> is <math>f</math>-saturated; that is, <math>R = f^{-1}(f(R)).</math></li>
</ol>

''Sufficient conditions for equality'':  Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>f</math> is injective.</li>
<li><math>R</math> is <math>f</math>-saturated; that is, <math>R = f^{-1}(f(R)).</math></li>
</ol>

=====Conditions for f(L∆R) = f(L)∆f(R){{anchor|Image of a symmetric difference ∆|Image of a symmetric difference}}=====

<math display=block>f\left(L ~\triangle~ R\right) ~\supseteq~ f(L) ~\triangle~ f(R) \qquad\qquad \text{ always holds}</math>

''Characterizations of equality'': The following statements are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f\left(L ~\triangle~ R\right) = f(L) ~\triangle~ f(R)</math></li>
<li><math>f\left(L ~\triangle~ R\right) \subseteq f(L) ~\triangle~ f(R)</math></li>
<li><math>f(L \,\setminus\, R) = f(L) \,\setminus\, f(R)</math> &nbsp;and&nbsp; <math>f(R \,\setminus\, L) = f(R) \,\setminus\, f(L)</math></li>
<li><math>f(L \,\setminus\, R) \subseteq f(L) \,\setminus\, f(R)</math> &nbsp;and&nbsp; <math>f(R \,\setminus\, L) \subseteq f(R) \,\setminus\, f(L)</math></li>
<li><math>L \cap f^{-1}(f(R)) ~\subseteq~ R</math> &nbsp;and&nbsp; <math>R \cap f^{-1}(f(L)) ~\subseteq~ L</math>
* The inclusions <math>L \cap f^{-1}(f(R)) ~\subseteq~ f^{-1}(f(L))</math> and <math>R \cap f^{-1}(f(L)) ~\subseteq~ f^{-1}(f(R))</math> always hold.</li>
<li><math>L \cap f^{-1}(f(R)) ~=~ R \cap f^{-1}(f(L))</math>
* If this above set equality holds, then this set will also be equal to both <math>L \cap R \cap \operatorname{domain} f</math> and <math>L \cap R \cap f^{-1}(f(L \cap R)).</math></li>
<li><math>L \cap f^{-1}(f(L \cap R)) ~=~ R \cap f^{-1}(f(L \cap R))</math> &nbsp;and&nbsp; <math>f(L \cap R) ~\supseteq~ f(L) \cap f(R).</math></li>
</ol>

''Necessary conditions for equality'' (excluding characterizations): If equality holds then the following are necessarily true:
<ol style="list-style-type: lower-latin;">
<li><math>f(L \cap R) = f(L) \cap f(R),</math> or equivalently <math>f(L \cap R) \supseteq f(L) \cap f(R).</math></li>
<li><math>L \cap f^{-1}(f(L \cap R)) ~=~ R \cap f^{-1}(f(L \cap R))</math></li>
</ol>

''Sufficient conditions for equality'': Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>f</math> is injective.</li>
<li>The restriction <math>f\big\vert_{L \cup R}</math> is injective.</li>
</ol>

====Exact formulas/equalities for images of set operations====

=====Formulas for f(L\R) ={{anchor|Image of set subtraction|Formulas for the image of set subtraction}}=====

For any function <math>f : X \to Y</math> and any sets <math>L</math> and <math>R,</math><ref group=proof name=ProofImagesOfSetSubtractions>Let <math>P := \left\{y \in Y : L \cap f^{-1}(y) \subseteq R\right\}</math> and let <math>(\star)</math> denote the set equality <math>f(L \setminus R) = Y \setminus P,</math> which will now be proven. If <math>y \in Y \setminus P</math> then <math>L \cap f^{-1}(y) \not\subseteq R</math> so there exists some <math>x \in L \cap f^{-1}(y) \setminus R;</math> now <math>f^{-1}(y) \subseteq X</math> implies <math>x \in L \cap X \setminus R</math> so that <math>y = f(x) \in f(L \cap X \setminus R) = f(L \setminus R).</math> To prove the reverse inclusion <math>f(L \setminus R) \subseteq Y \setminus P,</math> let <math>y \in f(L \setminus R)</math> so that there exists some <math>x \in X \cap L \setminus R</math> such that <math>y = f(x).</math> Then <math>x \in L \cap f^{-1}(y) \setminus R</math> so that <math>L \cap f^{-1}(y) \not\subseteq R</math> and thus <math>y \not\in P,</math> which proves that <math>y \in Y \setminus P,</math> as desired. <math>\blacksquare</math> Defining <math>Q := f(L) \cap P = \left\{ y \in f(L) : L \cap f^{-1}(y) \subseteq R \right\},</math> the identity <math>f(L \setminus R) = f(L) \setminus Q</math> follows from <math>(\star)</math> and the inclusions <math>f(L \setminus R) \subseteq f(L) \subseteq Y.</math> <math>\blacksquare</math></ref>
<math display=block>\begin{alignat}{4}
f(L \setminus R)
&= Y ~~~\;\,\, \setminus \left\{ y \in Y ~~~~~~~~~~\;\,   ~:~ L \cap f^{-1}(y) \subseteq R \right\} \\[0.4ex]
&= f(L)        \setminus \left\{ y \in f(L)~~~~~~~\,      ~:~ L \cap f^{-1}(y) \subseteq R \right\} \\[0.4ex]
&= f(L)        \setminus \left\{ y \in f(L \cap R)        ~:~ L \cap f^{-1}(y) \subseteq R \right\} \\[0.4ex]
&= f(L)        \setminus \left\{ y \in V~~~~~~~~~~~~\,    ~:~ L \cap f^{-1}(y) \subseteq R \right\} \qquad && \text{ for any superset } \quad V \supseteq f(L \cap R) \\[0.4ex]
&= f(S)        \setminus \left\{ y \in f(S)~~~~~~~\,      ~:~ L \cap f^{-1}(y) \subseteq R \right\} \qquad && \text{ for any superset } \quad S \supseteq L \cap X. \\[0.7ex]
\end{alignat}</math>

=====Formulas for f(X\R) ={{anchor|Image of set subtraction from the domain|Formulas for the image of set subtraction from the domain}}=====

Taking <math>L := X = \operatorname{domain} f</math> in the above formulas gives:
<math display=block>\begin{alignat}{4}
f(X \setminus R)
&= Y ~~~\;\,\,            \setminus \left\{ y \in Y ~~~~\;\,\,           :~ f^{-1}(y) \subseteq R \right\} \\[0.4ex]
&= f(X)                   \setminus \left\{ y \in f(X) ~                 :~ f^{-1}(y) \subseteq R \right\} \\[0.4ex]
&= f(X)                   \setminus \left\{ y \in f(R) ~                 :~ f^{-1}(y) \subseteq R \right\} \\[0.4ex]
&= f(X)                   \setminus \left\{ y \in W ~~~\;\,\,            :~ f^{-1}(y) \subseteq R \right\} \qquad \text{ for any superset } \quad W \supseteq f(R) \\[0.4ex]
\end{alignat}</math>
where the set <math>\left\{ y \in f(R) : f^{-1}(y) \subseteq R \right\}</math> is equal to the image under <math>f</math> of the largest <math>f</math>-saturated subset of <math>R.</math>
<ul>
<li>In general, only <math>f(X \setminus R) \,\supseteq\, f(X) \setminus f(R)</math> always holds and equality is not guaranteed; but replacing "<math>f(R)</math>" with its subset "<math>\left\{ y \in f(R) : f^{-1}(y) \subseteq R \right\}</math>" results in a formula in which equality is {{em|always}} guaranteed: 
<math display=block>f(X \setminus R) \,=\, f(X) \setminus  \left\{ y \in f(R) : f^{-1}(y) \subseteq R \right\}.</math> 
From this it follows that:<ref group=proof name=proofCharOfPullingLminusR>Let <math>f_R := \left\{ y \in f(L) : L \cap f^{-1}(y) \subseteq R \right\}</math> where because <math>f_R \subseteq f(R \cap L),</math> <math>f_R</math> is also equal to <math>f_R = \left\{ y \in f(R \cap L) : L \cap f^{-1}(y) \subseteq R \right\}.</math> As proved above, <math>f(L \setminus R) = f(L) \setminus f_R</math> so that <math>f(L) \setminus f(R) = f(L \setminus R)</math> if and only if <math>f(L) \setminus f(R) = f(L) \setminus f_R.</math> Since <math>f(L) \setminus f(R) = f(L) \setminus (f(L) \cap f(R)),</math> this happens if and only if <math>f(L) \setminus (f(L) \cap f(R)) = f(L) \setminus f_R.</math> Because <math>f(L) \cap f(R) \text{ and } f_R</math> are both subsets of <math>f(L),</math> the condition on the right hand side happens if and only if <math>f(L) \cap f(R) = f_R.</math> Because <math>f_R \subseteq f(R \cap L) \subseteq f(L) \cap f(R),</math> the equality <math>f(L) \cap f(R) = f_R</math> holds if and only if <math>f(L) \cap f(R) \subseteq f_R.</math> <math>\blacksquare</math> If <math>f(R) \subseteq f(L)</math> (such as when <math>L = X</math> or <math>R \subseteq L</math>) then <math>f(L) \cap f(R) \subseteq f_R</math> if and only if <math>f(R) \subseteq f_R.</math> In particular, taking <math>L = X</math> proves: <math>f(X \setminus R) = f(X) \setminus f(R)</math> if and only if <math>f(R) \subseteq \left\{ y \in f(R \cap X) : f^{-1}(y) \subseteq R \right\},</math> where <math>f(R \cap X) = f(R).</math> <math>\blacksquare</math></ref> 
<math display=block>f(X \setminus R) = f(X) \setminus f(R) \quad \text{ if and only if } \quad f(R) = \left\{ y \in f(R) : f^{-1}(y) \subseteq R \right\} \quad \text{ if and only if } \quad f^{-1}(f(R)) \subseteq R.</math>
</li>
<li>If <math>f_R := \left\{y \in f(X) : f^{-1}(y) \subseteq R\right\}</math> then <math>f(X \setminus R) = f(X) \setminus f_R,</math> which can be written more symmetrically as <math>f(X \setminus R) = f_X \setminus f_R</math> (since <math>f_X = f(X)</math>).
</li>
</ul>

=====Formulas for f(L∆R) ={{anchor|Image of symmetric difference|Formulas for the image of symmetric difference}}=====

It follows from <math>L \,\triangle\, R = (L \cup R) \setminus (L \cap R)</math> and the above formulas for the image of a set subtraction that for any function <math>f : X \to Y</math> and any sets <math>L</math> and <math>R,</math>
<math display=block>\begin{alignat}{4}
f(L \,\triangle\, R)
&= Y ~~~\;~~~\;~~~\; \setminus \left\{ y \in Y ~~~\,~~~\;~~~\,~   ~:~ L \cap f^{-1}(y) = R \cap f^{-1}(y)\right\} \\[0.4ex]
&= f(L \cup R) \setminus \left\{ y \in f(L \cup R)        ~:~ L \cap f^{-1}(y) = R \cap f^{-1}(y)\right\} \\[0.4ex]
&= f(L \cup R) \setminus \left\{ y \in f(L \cap R)        ~:~ L \cap f^{-1}(y) = R \cap f^{-1}(y)\right\} \\[0.4ex]
&= f(L \cup R) \setminus \left\{ y \in V ~~~\,~~~~~~~~~   ~:~ L \cap f^{-1}(y) = R \cap f^{-1}(y)\right\} \qquad && \text{ for any superset } \quad V \supseteq f(L \cap R) \\[0.4ex]
&= f(S) ~~\,~~~\,~\, \setminus \left\{ y \in f(S) ~~~\,~~~\;  ~:~ L \cap f^{-1}(y) = R \cap f^{-1}(y)\right\} \qquad && \text{ for any superset } \quad S \supseteq (L \cup R) \cap X. \\[0.7ex]
\end{alignat}</math>

=====Formulas for f(L) ={{anchor|Image of a set|Formulas for the image of a set}}=====

It follows from the above formulas for the image of a set subtraction that for any function <math>f : X \to Y</math> and any set <math>L,</math><math display=block>\begin{alignat}{4}
f(L)
&= Y ~~~\;\,           \setminus \left\{y \in Y ~~~\;\,           ~:~ f^{-1}(y) \cap L = \varnothing\right\} \\[0.4ex]
&= \operatorname{Im} f \setminus \left\{y \in \operatorname{Im} f ~:~ f^{-1}(y) \cap L = \varnothing\right\} \\[0.4ex]
&= W ~~~\,             \setminus \left\{y \in W ~~\;\,            ~:~ f^{-1}(y) \cap L = \varnothing\right\} \qquad \text{ for any superset } \quad W \supseteq f(L) \\[0.7ex]
\end{alignat}</math>

This is more easily seen as being a consequence of the fact that for any <math>y \in Y,</math> <math>f^{-1}(y) \cap L = \varnothing</math> if and only if <math>y \not\in f(L).</math>

=====Formulas for f(L⋂R) ={{anchor|Image of set intersection|Formulas for the image of set intersection}}=====

It follows from the above formulas for the image of a set that for any function <math>f : X \to Y</math> and any sets <math>L</math> and <math>R,</math>
<math display=block>\begin{alignat}{4}
f(L \cap R)
&= Y~~~~~ \setminus \left\{y \in Y~~~~~ ~:~ L \cap R \cap f^{-1}(y) = \varnothing\right\}        && \\[0.4ex]
&= f(L)   \setminus \left\{y \in f(L)   ~:~ L \cap R \cap f^{-1}(y) = \varnothing\right\}        && \\[0.4ex]
&= f(L)   \setminus \left\{y \in U~~~~~ ~:~ L \cap R \cap f^{-1}(y) = \varnothing\right\} \qquad && \text{ for any superset } \quad U \supseteq f(L) \\[0.4ex]
&= f(R)   \setminus \left\{y \in f(R)   ~:~ L \cap R \cap f^{-1}(y) = \varnothing\right\}        && \\[0.4ex]
&= f(R)   \setminus \left\{y \in V~~~~~ ~:~ L \cap R \cap f^{-1}(y) = \varnothing\right\} \qquad && \text{ for any superset } \quad V \supseteq f(R) \\[0.4ex]
&= f(L) \cap f(R) \setminus \left\{y \in f(L) \cap f(R) ~:~ L \cap R \cap f^{-1}(y) = \varnothing\right\}        && \\[0.7ex]
\end{alignat}</math>
where moreover, for any <math>y \in Y,</math> 
:<math>L \cap f^{-1}(y) \subseteq L \setminus R~</math> if and only if <math>~L \cap R \cap f^{-1}(y) = \varnothing~</math> if and only if <math>~R \cap f^{-1}(y) \subseteq R \setminus L~</math> if and only if <math>~y \not\in f(L \cap R).</math>
The sets <math>U</math> and <math>V</math> mentioned above could, in particular, be any of the sets <math>f(L \cup R), \;\operatorname{Im} f,</math> or <math>Y,</math> for example.

===(Pre)Images of set operations on (pre)images{{anchor|Images of preimages and preimages of images of set operations}}===

Let <math>L</math> and <math>R</math> be arbitrary sets, <math>f : X \to Y</math> be any map, and let <math>A \subseteq X</math> and <math>C \subseteq Y.</math> 

{| class="wikitable"
|-
! Image of preimage
! Preimage of image
! Additional assumptions on sets
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\left(f^{-1}(L) \cap R\right) = L \cap f(R)</math>{{sfn|Császár|1978|pp=102-120}}

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(f(L) \cap R) ~\supseteq~ L \cap f^{-1}(R)</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\left(f^{-1}(L) \cup R\right) ~=~ (L \cap \operatorname{Im} f) \cup f(R) ~\subseteq~ L \cup f(R)</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(f(A) \cup R) ~\supseteq~ A \cup f^{-1}(R)</math><ref name="lee-2010-p388">Lee p.388 of Lee, John M. (2010). Introduction to Topological Manifolds, 2nd Ed.</ref><br>
Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>f(A) \subseteq R.</math></li>
<li><math>A \subseteq f^{-1}(R).</math></li>
</ol>
|<math>A \subseteq X</math>
|}

'''(Pre)Images of operations on images'''

Since <math>f(L) \setminus f(L \setminus R) ~=~ \left\{y \in f(L \cap R) ~:~ L \cap f^{-1}(y) \subseteq R\right\},</math>

<math display=block>\begin{alignat}{4}
f^{-1}(f(L) \setminus f(L \setminus R)) 
&=&& f^{-1}\left(\left\{y \in f(L \cap R) ~:~ L \cap f^{-1}(y) \subseteq R\right\}\right) \\
&=&& \left\{x \in f^{-1}(f(L \cap R)) ~:~ L \cap f^{-1}(f(x)) \subseteq R\right\} \\
\end{alignat}</math>

Since <math>f(X) \setminus f(L \setminus R) ~=~ \left\{y \in f(X) ~:~ L \cap f^{-1}(y) \subseteq R\right\},</math>
<math display=block>\begin{alignat}{4}
f^{-1}(Y \setminus f(L \setminus R)) 
&~=~&& f^{-1}(f(X) \setminus f(L \setminus R)) \\
&=&& f^{-1}\left(\left\{y \in f(X) ~:~ L \cap f^{-1}(y) \subseteq R\right\}\right) \\
&=&& \left\{x \in X ~:~ L \cap f^{-1}(f(x)) \subseteq R\right\} \\
&~=~&& X \setminus f^{-1}(f(L \setminus R)) \\
\end{alignat}</math>

Using <math>L := X,</math> this becomes <math>~f(X) \setminus f(X \setminus R) ~=~ \left\{y \in f(R) ~:~ f^{-1}(y) \subseteq R\right\}~</math> and
<math display=block>\begin{alignat}{4}
f^{-1}(Y \setminus f(X \setminus R)) 
&~=~&& f^{-1}(f(X) \setminus f(X \setminus R)) \\
&=&& f^{-1}\left(\left\{y \in f(R) ~:~ f^{-1}(y) \subseteq R\right\}\right) \\
&=&& \left\{r \in R \cap X ~:~ f^{-1}(f(r)) \subseteq R\right\} \\
&\subseteq&& R \\
\end{alignat}</math>
and so
<math display=block>\begin{alignat}{4}
f^{-1}(Y \setminus f(L)) 
&~=~&& f^{-1}(f(X) \setminus f(L)) \\
&=&& f^{-1}\left(\left\{y \in f(X \setminus L) ~:~ f^{-1}(y) \cap L = \varnothing\right\}\right) \\
&=&& \{x \in X \setminus L ~:~ f(x) \not\in f(L)\} \\
&=&& X \setminus f^{-1}(f(L)) \\
&\subseteq&& X \setminus L \\
\end{alignat}</math>

===(Pre)Images and Cartesian products Π===

Let <math>\prod Y_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \prod_{j \in J} Y_j</math> and for every <math>k \in J,</math> let 
<math display=block>\pi_k ~:~ \prod_{j \in J} Y_j ~\to~ Y_k</math> 
denote the canonical projection onto <math>Y_k.</math> 

'''Definitions'''

Given a collection of maps <math>F_j : X \to Y_j</math> indexed by <math>j \in J,</math> define the map
<math display=block>\begin{alignat}{4}
\left(F_j\right)_{j \in J} :\;&& X &&\;\to\;    & \prod_{j \in J} Y_j \\[0.3ex]
                                          && x &&\;\mapsto\;& \left(F_j\left(x_j\right)\right)_{j \in J}, \\
\end{alignat}</math>
which is also denoted by <math>F_{\bull} = \left(F_j\right)_{j \in J}.</math> This is the unique map satisfying 
<math display=block>\pi_j \circ F_{\bull} = F_j \quad \text{ for all } j \in J.</math> 

Conversely, if given a map <math display=block>F ~:~ X ~\to~ \prod_{j \in J} Y_j</math> then <math>F = \left(\pi_j \circ F\right)_{j \in J}.</math>
Explicitly, what this means is that if 
<math display=block>F_k ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \pi_k \circ F ~:~ X ~\to~ Y_k</math> 
is defined for every <math>k \in J,</math> then <math>F</math> the unique map satisfying: <math>\pi_j \circ F = F_j</math> for all <math>j \in J;</math> or said more briefly, <math>F = \left(F_j\right)_{j \in J}.</math>

The map <math>F_{\bull} = \left(F_j\right)_{j \in J} ~:~ X ~\to~ \prod_{j \in J} Y_j</math> should not be confused with the [[Cartesian product]] <math>\prod_{j \in J} F_j</math> of these maps, which is by definition is the map 
<math display=block>\begin{alignat}{4}
\prod_{j \in J} F_j :\;&& \prod_{j \in J} X          &&~\;\to\;~    & \prod_{j \in J} Y_j \\[0.3ex]
                       && \left(x_j\right)_{j \in J} &&~\;\mapsto\;~& \left(F_j\left(x_j\right)\right)_{j \in J} \\
\end{alignat}</math>
with domain <math>\prod_{j \in J} X = X^J</math> rather than <math>X.</math>

'''Preimage and images of a Cartesian product'''

Suppose <math>F_{\bull} = \left(F_j\right)_{j \in J} ~:~ X ~\to~ \prod_{j \in J} Y_j.</math> 

If <math>A ~\subseteq~ X</math> then
<math display=block>F_{\bull}(A) ~~\;\color{Red}{\subseteq}\color{Black}{}\;~~ \prod_{j \in J} F_j(A).</math> 

If <math>B ~\subseteq~ \prod_{j \in J} Y_j</math> then 
<math display=block>F_{\bull}^{-1}(B) ~~\;\color{Red}{\subseteq}\color{Black}{}\;~~ \bigcap_{j \in J} F_j^{-1}\left(\pi_j(B)\right)</math> 
where equality will hold if <math>B = \prod_{j \in J} \pi_j(B),</math> in which case <math display=inline>F_{\bull}^{-1}(B) = \displaystyle\bigcap_{j \in J} F_j^{-1}\left(\pi_j(B)\right)</math> and 
{{NumBlk||<math display=block>F_{\bull}^{-1}\left(\prod_{j \in J} \pi_j(B)\right) ~=~ \bigcap_{j \in J} F_j^{-1}\left(\pi_j(B)\right).</math>|{{EquationRef|Eq. 11a}}}}

For equality to hold, it suffices for there to exist a family <math>\left(B_j\right)_{j \in J}</math> of subsets <math>B_j \subseteq Y_j</math> such that <math>B = \prod_{j \in J} B_j,</math> in which case: 
{{NumBlk||<math display=block>F_{\bull}^{-1}\left(\prod_{j \in J} B_j\right) ~=~ \bigcap_{j \in J} F_j^{-1}\left(B_j\right)</math>|{{EquationRef|Eq. 11b}}}}
and <math>\pi_j(B) = B_j</math> for all <math>j \in J.</math>

===(Pre)Image of a single set===

{| class="wikitable"
|-
! Image
! Preimage
! Additional assumptions
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
f(L) &= f(L \cap \operatorname{domain} f) \\
&= f(L \cap X) \\
&= Y ~~~~\,            \setminus \left\{ y \in Y ~~~~\,            : f^{-1}(y) \subseteq X \setminus L \right\} \\
&= \operatorname{Im} f \setminus \left\{ y \in \operatorname{Im} f : f^{-1}(y) \subseteq X \setminus L \right\} \\
\end{alignat}</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
f^{-1}(L) 
&= f^{-1}(L \cap \operatorname{Im} f) \\
&= f^{-1}(L \cap Y)
\end{alignat}</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f(X) = \operatorname{Im} f \subseteq Y</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
f^{-1}(Y)                   &= X \\
f^{-1}(\operatorname{Im} f) &= X
\end{alignat}</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
f(L) 
&= f(L \cap R  ~&&\cup~  &&(&&L \setminus R)) \\
&= f(L \cap R) ~&&\cup~ f&&(&&L \setminus R)
\end{alignat}</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
f^{-1}(L) 
&= f^{-1}(L \cap R  &&\cup       &&( &&L                           &&\setminus  &&R))  \\
&= f^{-1}(L \cap R) &&\cup f^{-1}&&( &&L                           &&\setminus  &&R) \\
&= f^{-1}(L \cap R) &&\cup f^{-1}&&( &&L                           &&\setminus [&&R \cap \operatorname{Im} f]) \\
&= f^{-1}(L \cap R) &&\cup f^{-1}&&([&&L \cap \operatorname{Im} f] &&\setminus  &&R) \\
&= f^{-1}(L \cap R) &&\cup f^{-1}&&([&&L \cap \operatorname{Im} f] &&\setminus [&&R \cap \operatorname{Im} f]) 
\end{alignat}</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\operatorname{Im} f  = f(X) ~=~ f(L) \cup f(X \setminus L)</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\begin{alignat}{4}
X 
&= f^{-1}(L) \cup f^{-1}(Y                   &&\setminus L) \\
&= f^{-1}(L) \cup f^{-1}(\operatorname{Im} f &&\setminus L)
\end{alignat}</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\big\vert_L(R) = f(L \cap R)</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>\left(f\big\vert_L\right)^{-1}(R) = L \cap f^{-1}(R)</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>(g \circ f)(L) ~=~ g(f(L))</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>(g \circ f)^{-1}(L) ~=~ f^{-1}\left(g^{-1}(L)\right)</math>
|None (<math>f</math> and <math>g</math> are arbitrary functions).
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\left(f^{-1}(L)\right) = L \cap \operatorname{Im} f</math><br>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(f(L)) ~\supseteq~ L \cap f^{-1}(\operatorname{Im} f) = L \cap f^{-1}(Y)</math>{{sfn|Császár|1978|pp=102-120}}
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\left(f^{-1}(Y)\right) = f\left(f^{-1}(\operatorname{Im} f)\right) = f(X)</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(f(X)) = f^{-1}(\operatorname{Im} f) = X</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\left(f^{-1}(f(L))\right) = f(L)</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}\left(f\left(f^{-1}(L)\right)\right) = f^{-1}(L)</math>
|None
|}

====Containments ⊆ and intersections ⋂ of images and preimages====

'''Equivalences and implications of images and preimages'''

{| class="wikitable"
|-
! Image
! Preimage
! Additional assumptions on sets
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f(L) \subseteq \operatorname{Im} f \subseteq Y</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(L) = X</math> if and only if <math>\operatorname{Im} f \subseteq L</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f(L) = \varnothing</math> if and only if <math>L \cap \operatorname{domain} f = \varnothing</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(L) = \varnothing</math> if and only if <math>L \cap \operatorname{Im} f = \varnothing</math>
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f(A) = \varnothing</math> if and only if <math>A = \varnothing</math>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(C) = \varnothing</math> if and only if <math>C \subseteq Y \setminus \operatorname{Im} f</math>
|<math>A \subseteq X</math> and <math>C \subseteq Y</math>
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>L \subseteq R</math> implies <math>f(L) \subseteq f(R)</math>{{sfn|Császár|1978|pp=102-120}}

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>L \subseteq R</math> implies <math>f^{-1}(L) \subseteq f^{-1}(R)</math>{{sfn|Császár|1978|pp=102-120}}
|None
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|The following are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f(L) \subseteq f(R)</math></li>
<li><math>f(L \cup R) = f(R)</math></li>
<li><math>L \cap \operatorname{domain} f ~\subseteq~ f^{-1}(f(R))</math></li>
</ol>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|The following are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f^{-1}(L) \subseteq f^{-1}(R)</math></li>
<li><math>f^{-1}(L \cup R) = f^{-1}(R)</math></li>
<li><math>L \cap \operatorname{Im} f \subseteq R</math></li>
</ol>
If <math>C ~\subseteq~ \operatorname{Im} f</math> then <math>f^{-1}(C) ~\subseteq~ f^{-1}(R)</math> if and only if <math>C ~\subseteq~ R.</math>
|<math>C \subseteq \operatorname{Im} f</math>
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|The following are equivalent when <math>C \subseteq Y:</math>
<ol style="list-style-type: lower-latin;">
<li><math>C \subseteq f(R)</math></li>
<li><math>f(B) = C</math> for some <math>B \subseteq R</math></li>
<li>{{nowrap|<math>f(B) = C</math> for some <math>B \subseteq R \cap \operatorname{domain} f</math>}}</li>
</ol>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|The following are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>L \subseteq f^{-1}(R)</math></li>
<li><math>f(L) \subseteq R</math> and <math>L \subseteq \operatorname{domain} f</math></li>
</ol>
The following are equivalent when <math>A \subseteq X:</math>
<ol style="list-style-type: lower-latin;">
<li><math>A \subseteq f^{-1}(R)</math></li>
<li><math>f(A) \subseteq R</math></li>
</ol>
|<math>A \subseteq X</math> and <math>C \subseteq Y</math>
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|The following are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f(A) ~\supseteq~ f(X \setminus A)</math></li>
<li><math>f(A) = \operatorname{Im} f</math></li>
</ol>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|The following are equivalent:
<ol style="list-style-type: lower-latin;">
<li><math>f^{-1}(C) ~\supseteq~ f^{-1}(Y \setminus C)</math></li>
<li><math>f^{-1}(C) = X</math></li>
</ol>
|<math>A \subseteq X</math> and <math>C \subseteq Y</math>
|-
|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f\left(f^{-1}(L)\right) \subseteq L</math>{{sfn|Császár|1978|pp=102-120}}<br>
Equality holds if {{em|and only if}} the following is true:
<ol>
<li><math>L \subseteq \operatorname{Im} f.</math><ref name="halmos-1960-p39">Lee {{harvnb|Halmos|1960|p=39}}</ref><ref name="munkres-2000-p19">Lee {{harvnb|Munkres|2000|p=19}}</ref></li>
</ol>
Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>L \subseteq Y</math> and <math>f : X \to Y</math> is surjective.</li>
</ol>

|style="padding: 0.5em 0.8em 0.5em 1.5em;"|<math>f^{-1}(f(A)) \supseteq A</math><br>
Equality holds if {{em|and only if}} the following is true:
<ol>
<li><math>A</math> is <math>f</math>-saturated.</li>
</ol>
Equality holds if any of the following are true:
<ol style="list-style-type: lower-latin;">
<li><math>f</math> is injective.<ref name="halmos-1960-p39"/><ref name="munkres-2000-p19" /></li>
</ol>
|<math>A \subseteq X</math>
|}

'''Intersection of a set and a (pre)image'''

The following statements are equivalent:
<ol>
<li><math>\varnothing = f(L) \cap R</math></li>
<li><math>\varnothing = L \cap f^{-1}(R)</math>{{sfn|Császár|1978|pp=102-120}}</li>
<li><math>\varnothing = f^{-1}(f(L)) \cap f^{-1}(R)</math></li>
<li><math>\varnothing = f^{-1}(f(L) \cap R)</math></li>
</ol>

Thus for any <math>t,</math>{{sfn|Császár|1978|pp=102-120}} <math display=block>t \not\in f(L) \quad \text{ if and only if } \quad L \cap f^{-1}(t) = \varnothing.</math>

==Sequences and collections of families of sets==

===Definitions===

A {{em|[[Family of sets|family of sets]]}} or simply a {{em|family}} is a set whose elements are sets.  
A {{em|family over <math>X</math>}} is a family of subsets of <math>X.</math> 

The {{em|[[Power set|power set]]}} of a set <math>X</math> is the set of all subsets of <math>X</math>: 
<math display=block>\wp(X) ~\colon=~ \{\; S ~:~ S \subseteq X\; \}.</math>

'''Notation for sequences of sets'''

Throughout, <math>S \text{ and } T</math> will be arbitrary sets and <math>S_{\bull}</math> and will denote a [[Net (mathematics)|net]] or a [[Sequence|sequence]] of sets where if it is a sequence then this will be indicated by either of the notations 
<math display=block>S_{\bull} = \left(S_i\right)_{i=1}^{\infty} \qquad \text{ or } \qquad S_{\bull} = \left(S_i\right)_{i \in \N}</math>
where <math>\N</math> denotes the natural numbers. 
A notation <math>S_{\bull} = \left(S_i\right)_{i \in I}</math> indicates that <math>S_{\bull}</math> is a [[Net (mathematics)|net]] [[Directed set|directed]] by <math>(I, \leq),</math> which (by definition) is a [[Sequence|sequence]] if the set <math>I,</math> which is called the net's indexing set, is the natural numbers (that is, if <math>I = \N</math>) and <math>\,\leq\,</math> is the natural order on <math>\N.</math> 

'''Disjoint and monotone sequences of sets'''

If <math>S_i \cap S_j = \varnothing</math> for all distinct indices <math>i \neq j</math> then <math>S_{\bull}</math> is called a {{em|pairwise disjoint}} or simply a {{em|disjoint}}.  
A sequence or net <math>S_{\bull}</math> of set is called {{em|increasing}} or {{em|non-decreasing}} if (resp. {{em|decreasing}} or {{em|non-increasing}}) if for all indices <math>i \leq j,</math> <math>S_i \subseteq S_j</math> (resp. <math>S_i \supseteq S_j</math>). 
A sequence or net <math>S_{\bull}</math> of set is called {{em|strictly increasing}} (resp. {{em|strictly decreasing}}) if it is non-decreasing (resp. is non-increasing) and also <math>S_i \neq S_j</math> for all {{em|distinct}} indices <math>i \text{ and } j.</math> 
It is called {{em|monotone}} if it is non-decreasing or non-increasing and it is called {{em|strictly monotone}} if it is strictly increasing or strictly decreasing. 

A sequences or net <math>S_{\bull}</math> is said to {{em|increase to <math>S,</math>}} denoted by <math>S_{\bull} \uparrow S</math>{{sfn|Durrett|2019|pp=1-8}} or <math>S_{\bull} \nearrow S,</math> if <math>S_{\bull}</math> is increasing and the union of all <math>S_i</math> is <math>S;</math> that is, if <math display=block>\bigcup_{n} S_n = S \qquad \text{ and } \qquad S_i \subseteq S_j \quad \text{ whenever } i \leq j.</math>
It is said to {{em|decrease to <math>S,</math>}} denoted by <math>S_{\bull} \downarrow S</math>{{sfn|Durrett|2019|pp=1-8}} or <math>S_{\bull} \searrow S,</math> if <math>S_{\bull}</math> is increasing and the intersection of all <math>S_i</math> is <math>S</math> that is, if <math display=block>\bigcap_{n} S_n = S \qquad \text{ and } \qquad S_i \supseteq S_j \quad \text{ whenever } i \leq j.</math> 

'''Definitions of elementwise operations on families'''

If <math>\mathcal{L} \text{ and } \mathcal{R}</math> are families of sets and if <math>S</math> is any set then define:{{sfn|Császár|1978|pp=53-65}}  
<math display=block>\mathcal{L} \;(\cup)\; \mathcal{R} ~\colon=~ \{~ L \cup R ~:~ L \in \mathcal{L} ~\text{ and }~ R \in \mathcal{R} ~\}</math>
<math display=block>\mathcal{L} \;(\cap)\; \mathcal{R} ~\colon=~ \{~ L \cap R ~:~ L \in \mathcal{L} ~\text{ and }~ R \in \mathcal{R} ~\}</math>
<math display=block>\mathcal{L} \;(\setminus)\; \mathcal{R} ~\colon=~ \{~ L \setminus R ~:~ L \in \mathcal{L} ~\text{ and }~ R \in \mathcal{R} ~\}</math>
<math display=block>\mathcal{L} \;(\triangle)\; \mathcal{R} ~\colon=~ \{~ L \;\triangle\; R ~:~ L \in \mathcal{L} ~\text{ and }~ R \in \mathcal{R} ~\}</math>
<math display=block>\mathcal{L}\big\vert_S ~\colon=~ \{ L \cap S ~:~ L \in \mathcal{L} \} = \mathcal{L} \;(\cap)\; \{S\}</math>
which are respectively called '''{{em|elementwise}} {{em|union}}''', '''{{em|elementwise}} {{em|intersection}}''', '''{{em|elementwise}}''' ('''{{em|set}}''') '''{{em|difference}}''', '''{{em|elementwise}} {{em|symmetric difference}}''', and the '''{{em|trace}}'''/'''{{em|restriction of <math>\mathcal{L}</math> to <math>S.</math>}}''' The regular union, intersection, and set difference are all defined as usual and are denoted with their usual notation: <math>\mathcal{L} \cup \mathcal{R}, \mathcal{L} \cap \mathcal{R}, \mathcal{L} \;\triangle\; \mathcal{R},</math> and <math>\mathcal{L} \setminus \mathcal{R},</math> respectively.  
These elementwise operations on families of sets play an important role in, among other subjects, the theory of [[Filter (set theory)|filters]] and prefilters on sets. 

The {{em|upward closure in <math>X</math>}} of a family <math>\mathcal{L} \subseteq \wp(X)</math> is the family: 
<math display=block>\mathcal{L}^{\uparrow X} ~\colon=~ \bigcup_{L \in \mathcal{L}} \{\; S ~:~ L \subseteq S \subseteq X \;\} ~=~ \{\; S \subseteq X ~:~ \text{ there exists } L \in \mathcal{L} \text{ such that } L \subseteq S \;\}</math> 
and the {{em|downward closure of <math>\mathcal{L}</math>}} is the family:
<math display=block>\mathcal{L}^{\downarrow} ~\colon=~ \bigcup_{L \in \mathcal{L}} \wp(L) ~=~ \{\; S ~:~ \text{ there exists } L \in \mathcal{L} \text{ such that } S \subseteq L \;\}.</math> 

====Definitions of categories of families of sets====


A family <math>\mathcal{L}</math> is called {{em|isotone}}, {{em|ascending}}, or {{em|upward closed}} in <math>X</math> if <math>\mathcal{L} \subseteq \wp(X)</math> and <math>\mathcal{L} = \mathcal{L}^{\uparrow X}.</math>{{sfn|Császár|1978|pp=53-65}}  
A family <math>\mathcal{L}</math> is called {{em|downward closed}} if <math>\mathcal{L} = \mathcal{L}^{\downarrow}.</math>

A family <math>\mathcal{L}</math> is said to be: 
* {{em|closed under finite intersections}} (resp. {{em|closed under finite unions}}) if whenever <math>L, R \in \mathcal{L}</math> then <math>L \cap R \in \mathcal{L}</math> (respectively, <math>L \cup R \in \mathcal{L}</math>). 
* {{em|closed under [[Countable set|countable]] intersections}} (resp. {{em|closed under countable unions}}) if whenever <math>L_1, L_2, L_3, \ldots</math> are elements of <math>\mathcal{L}</math> then so is their intersections <math>\bigcap_{i=1}^{\infty} L_i := L_1 \cap L_2 \cap L_3 \cap \cdots</math> (resp. so is their union <math>\bigcup_{i=1}^{\infty} L_i := L_1 \cup L_2 \cup L_3 \cup \cdots</math>). 
* {{em|closed under [[Complement (set theory)|complementation]] in}} (or {{em|with respect to}}) {{em|<math>X</math>}} if whenever <math>L \in \mathcal{L}</math> then <math>X \setminus L \in \mathcal{L}.</math> 

A family <math>\mathcal{L}</math> of sets is called a/an: 
* {{em|[[Pi-system|{{pi}}−system]]}} if <math>\mathcal{L} \neq \varnothing</math> and <math>\mathcal{L}</math> is closed under finite-intersections. 
** Every non-empty family <math>\mathcal{L}</math> is contained in a unique smallest (with respect to <math>\subseteq</math>) {{pi}}−system that is denoted by <math>\pi(\mathcal{L})</math> and called {{em|the {{pi}}−system generated by <math>\mathcal{L}.</math>}} 
* {{em|filter subbase}} and is said to have {{em|the [[Finite intersection property|finite intersection property]]}} if <math>\mathcal{L} \neq \varnothing</math> and <math>\varnothing \not\in \pi(\mathcal{L}).</math> 
* {{em|[[Filter (set theory)|filter]] on <math>X</math>}} if <math>\mathcal{L} \neq \varnothing</math> is a family of subsets of <math>X</math> that is a {{pi}}−system, is upward closed in <math>X,</math> and is also {{em|proper}}, which by definition means that it does not contain the empty set as an element. 
* {{em|prefilter}} or {{em|filter base}} if it is a non-empty family of subsets of some set <math>X</math> whose upward closure in <math>X</math> is a filter on <math>X.</math> 
* {{em|[[Field of sets|algebra on <math>X</math>]]}} is a non-empty family of subsets of <math>X</math> that contains the empty set, forms a {{pi}}−system, and is also closed under complementation with respect to <math>X.</math> 
* {{em|[[Σ-algebra|σ-algebra]] on <math>X</math>}} is an algebra on <math>X</math> that is closed under countable unions (or equivalently, closed under countable intersections).

[[Sequence]]s of sets often arise in measure theory.

'''Algebra of sets'''


A [[Family of sets|family]] <math>\Phi</math> of subsets of a set <math>X</math> is said to be '''{{em|an algebra of sets}}''' if <math>\varnothing \in \Phi</math> and for all <math>L, R \in \Phi,</math> all three of the sets <math>X \setminus R, \,L \cap R,</math> and <math>L \cup R</math> are elements of <math>\Phi.</math><ref name="Encyclopedia of math">{{cite web|url=https://encyclopediaofmath.org/wiki/Algebra_of_sets|title=Algebra of sets|website=Encyclopediaofmath.org|date=16 August 2013|access-date=8 November 2020}}</ref> 
The [[Algebra of sets|article on this topic]] lists set identities and other relationships these three operations.

Every algebra of sets is also a [[Ring of sets|ring of sets]]<ref name="Encyclopedia of math"/> and a [[Pi-system|π-system]]. 

'''Algebra generated by a family of sets'''

Given any family <math>\mathcal{S}</math> of subsets of <math>X,</math> there is a unique smallest<ref group=note>Here "smallest" means relative to subset containment. So if <math>\Phi</math> is any algebra of sets containing <math>\mathcal{S},</math> then <math>\Phi_{\mathcal{S}} \subseteq \Phi.</math></ref> algebra of sets in <math>X</math> containing <math>\mathcal{S}.</math><ref name="Encyclopedia of math"/> 
It is called '''{{em|the algebra generated by <math>\mathcal{S}</math>}}''' and it will be denote it by <math>\Phi_{\mathcal{S}}.</math> 
This algebra can be constructed as follows:<ref name="Encyclopedia of math"/>

<ol>
<li>If <math>\mathcal{S} = \varnothing</math> then <math>\Phi_{\mathcal{S}} = \{\varnothing, X\}</math> and we are done. Alternatively, if <math>\mathcal{S}</math> is empty then <math>\mathcal{S}</math> may be replaced with <math>\{\varnothing\}, \{X\}, \text{ or } \{\varnothing, X\}</math> and continue with the construction.</li>
<li>Let <math>\mathcal{S}_0</math> be the family of all sets in <math>\mathcal{S}</math> together with their complements (taken in <math>X</math>).</li>
<li>Let <math>\mathcal{S}_1</math> be the family of all possible finite intersections of sets in <math>\mathcal{S}_0.</math><ref group=note>Since <math>\mathcal{S} \neq \varnothing,</math> there is some <math>S \in \mathcal{S}_0</math> such that its complement also belongs to <math>\mathcal{S}_0.</math> The intersection of these two sets implies that <math>\varnothing \in \mathcal{S}_1.</math> The union of these two sets is equal to <math>X,</math> which implies that <math>X \in \Phi_{\mathcal{S}}.</math></ref></li>
<li>Then the algebra generated by <math>\mathcal{S}</math> is the set <math>\Phi_{\mathcal{S}}</math> consisting of all possible finite unions of sets in <math>\mathcal{S}_1.</math></li>
</ol>

====Elementwise operations on families====

Let <math>\mathcal{L}, \mathcal{M},</math> and <math>\mathcal{R}</math> be families of sets over <math>X.</math> 
On the left hand sides of the following identities, <math>\mathcal{L}</math> is the {{em|L}}{{hsp}}eft most family, <math>\mathcal{M}</math> is in the {{em|M}}{{hsp}}iddle, and <math>\mathcal{R}</math> is the {{em|R}}{{hsp}}ight most set. 

'''Commutativity''':{{sfn|Császár|1978|pp=53-65}}  
<math display=block>\mathcal{L} \;(\cup)\; \mathcal{R} = \mathcal{R} \;(\cup)\; \mathcal{L}</math>
<math display=block>\mathcal{L} \;(\cap)\; \mathcal{R} = \mathcal{R} \;(\cap)\; \mathcal{L}</math>

'''Associativity''':{{sfn|Császár|1978|pp=53-65}}  
<math display=block>[\mathcal{L} \;(\cup)\; \mathcal{M}] \;(\cup)\; \mathcal{R} = \mathcal{L} \;(\cup)\; [\mathcal{M} \;(\cup)\; \mathcal{R}]</math>
<math display=block>[\mathcal{L} \;(\cap)\; \mathcal{M}] \;(\cap)\; \mathcal{R} = \mathcal{L} \;(\cap)\; [\mathcal{M} \;(\cap)\; \mathcal{R}]</math>

'''Identity''':
<math display=block>\mathcal{L} \;(\cup)\; \{\varnothing\} = \mathcal{L}</math>
<math display=block>\mathcal{L} \;(\cap)\; \{X\} = \mathcal{L}</math>
<math display=block>\mathcal{L} \;(\setminus)\; \{\varnothing\} = \mathcal{L}</math>

'''Domination''':
<math display=block>\mathcal{L} \;(\cup)\; \{X\} = \{X\} ~~~~\text{ if } \mathcal{L} \neq \varnothing</math>
<math display=block>\mathcal{L} \;(\cap)\; \{\varnothing\} = \{\varnothing\} ~~~~\text{ if } \mathcal{L} \neq \varnothing</math>
<math display=block>\mathcal{L} \;(\cup)\; \varnothing = \varnothing</math>
<math display=block>\mathcal{L} \;(\cap)\; \varnothing = \varnothing</math>
<math display=block>\mathcal{L} \;(\setminus)\; \varnothing = \varnothing</math>
<math display=block>\varnothing \;(\setminus)\; \mathcal{R} = \varnothing</math>

===Power set===

<math display=block>\wp(L \cap R) ~=~ \wp(L) \cap \wp(R)</math>
<math display=block>\wp(L \cup R) ~=~ \wp(L) \ (\cup)\  \wp(R) ~\supseteq~ \wp(L) \cup \wp(R).</math>

If <math>L</math> and <math>R</math> are subsets of a vector space <math>X</math> and if <math>s</math> is a scalar then
<math display=block>\wp(s L) ~=~ s \wp(L)</math>
<math display=block>\wp(L + R) ~\supseteq~ \wp(L) + \wp(R).</math>

===Sequences of sets===

Suppose that <math>L</math> is any set such that <math>L \supseteq R_i</math> for every index <math>i.</math> 
If <math>R_\bull</math> decreases to <math>R</math> then <math>L \setminus R_\bull := \left(L \setminus R_i\right)_i</math> increases to <math>L \setminus R</math>{{sfn|Durrett|2019|pp=1-8}} 
whereas if instead <math>R_\bull</math> increases to <math>R</math> then <math>L \setminus R_\bull</math> decreases to <math>L \setminus R.</math> 

If <math>L \text{ and } R</math> are arbitrary sets and if <math>L_\bull = \left(L_i\right)_i</math> increases (resp. decreases) to <math>L</math> then <math>\left(L_i \setminus R\right)_i</math> increase (resp. decreases) to <math>L \setminus R.</math> 

====Partitions====

Suppose that <math>S_\bull = \left(S_i\right)_{i = 1}^\infty</math> is any sequence of sets, that <math>S \subseteq \bigcup_i S_i</math> is any subset, and for every index <math>i,</math> let <math>D_i = \left(S_i \cap S\right) \setminus \bigcup_{m=1}^i \left(S_m \cap S\right).</math> 
Then <math>S = \bigcup_i D_i</math> and <math>D_\bull := \left(D_i\right)_{i=1}^\infty</math> is a sequence of pairwise disjoint sets.{{sfn|Durrett|2019|pp=1-8}} 

Suppose that <math>S_\bull = \left(S_i\right)_{i = 1}^\infty</math> is non-decreasing, let <math>S_0 = \varnothing,</math> and let <math>D_i = S_i \setminus S_{i-1}</math> for every <math>i = 1, 2, \ldots.</math> Then <math>\bigcup_i S_i = \bigcup_i D_i</math> and <math>D_\bull = \left(D_i\right)_{i=1}^\infty</math> is a sequence of pairwise disjoint sets.{{sfn|Durrett|2019|pp=1-8}}

==See also==

* {{annotated link|Algebra of sets}}
* {{annotated link|Complement (set theory)}}
* {{annotated link|Inclusion–exclusion principle}}
* {{annotated link|Intersection (set theory)}}
* {{annotated link|List of mathematical identities}}
* {{annotated link|Naive set theory}}
* {{annotated link|Pigeonhole principle}}
* {{annotated link|Set (mathematics)}}
* {{annotated link|Simple theorems in the algebra of sets}}
* {{annotated link|Union (set theory)}}

==Notes==

'''Notes'''

{{reflist|group=note}}

'''Proofs'''

{{reflist|group=proof}}

==Citations==

{{reflist}}

==References==
* {{cite book|last=Artin|first= Michael|title= Algebra|year=1991|publisher=Prentice Hall|isbn= 81-203-0871-9}}
* {{cite book|last=Blyth|first=T.S.|title=Lattices and Ordered Algebraic Structures|publisher= Springer|year= 2005|isbn=1-85233-905-5}}.
* {{cite journal|last=Bylinski|first=Czeslaw|date=2004|title=Some Basic Properties of Sets|url=http://www.mizar.org/JFM/Vol1/zfmisc_1.html|journal=Journal of Formalized Mathematics|volume=1|issue=|pages=|doi=|access-date=5 October 2021}}
* Courant, Richard, Herbert Robbins, Ian Stewart, ''What is mathematics?: An Elementary Approach to Ideas and Methods'', Oxford University Press US, 1996. {{ISBN|978-0-19-510519-3}}. [https://books.google.com/books?id=UfdossHPlkgC&dq=%22algebra+of+sets%22&pg=PA17-IA8 "SUPPLEMENT TO CHAPTER II THE ALGEBRA OF SETS"].
* {{Császár General Topology}} <!--{{sfn|Császár|1978|p=}}-->
* {{Dixmier General Topology}} <!--{{sfn|Dixmier|1984|p=}}-->
* {{Dolecki Mynard Convergence Foundations Of Topology}} <!--{{sfn|Dolecki|Mynard|2016|p=}}-->
* {{Dugundji Topology}} <!--{{sfn|Dugundji|1966|p=}}-->
* {{Durrett Probability Theory and Examples 5th Edition}} <!--{{sfn|Durrett|2019|p=}}-->
* {{cite book|last=Halmos|first=Paul R.|title=Naive set theory|url=https://archive.org/details/naivesettheory0000halm|url-access=registration|series=The University Series in Undergraduate Mathematics|publisher=van Nostrand Company|year=1960|isbn=9780442030643 |zbl=0087.04403}}
* {{Joshi Introduction to General Topology}} <!--{{sfn|Joshi|1983|p=}}-->
* {{cite book|last1=Kelley|first1=John L.|title=General Topology|edition=2|series=[[Graduate Texts in Mathematics]]|volume=27|year=1985|publisher=Birkhäuser|isbn=978-0-387-90125-1}}
* {{Köthe Topological Vector Spaces I}} <!--{{sfn|Köthe|1969|p=}}-->
* {{cite book|last=Monk|first=James Donald|title=Introduction to Set Theory|publisher=McGraw-Hill|publication-place=New York|url=http://euclid.colorado.edu/~monkd/monk11.pdf|year=1969|series=International series in pure and applied mathematics|isbn=978-0-07-042715-0|oclc=1102}} <!--{{sfn|Monk|1969|p=}}-->
* {{Munkres Topology|edition=2}} <!--{{sfn|Munkres|2000|p=}}-->
* {{Narici Beckenstein Topological Vector Spaces|edition=2}} <!--{{sfn|Narici|Beckenstein|2011|p=}}-->
* {{cite journal|last=Padlewska|first=Beata|date=1990|title=Families of Sets|url=http://mizar.org/JFM/Vol1/setfam_1.html|journal=Journal of Formalized Mathematics|volume=1|issue=|pages=1|doi=|access-date=5 October 2021}}
* {{Schechter Handbook of Analysis and Its Foundations}} <!--{{sfn|Schechter|1996|p=}}-->
* {{Schubert Topology}} <!--{{sfn|Schubert|1968|p=}}-->
* Stoll, Robert R.; {{em|Set Theory and Logic}}, Mineola, N.Y.: Dover Publications (1979) {{ISBN|0-486-63829-4}}. [https://books.google.com/books?id=3-nrPB7BQKMC&pg=PA16 "The Algebra of Sets", pp 16&mdash;23].
* {{cite journal|last=Trybulec|first=Zinaida|date=2002|title=Properties of subsets|url=https://fm.mizar.org/1990-1/pdf1-1/subset_1.pdf|journal=Journal of Formalized Mathematics|volume=1|issue=|pages=1|doi=|access-date=5 October 2021}}
* {{Wilansky Modern Methods in Topological Vector Spaces|edition=1}} <!--{{sfn|Wilansky|2013|p=}}-->
* {{Willard General Topology}} <!--{{sfn|Willard|2004|p=}}-->

==External links==

* [http://www.apronus.com/provenmath/btheorems.htm Operations on Sets at ProvenMath]

{{Mathematical logic}}

{{DEFAULTSORT:Set identities and relations}}
[[Category:Articles containing proofs]]
[[Category:Basic concepts in infinite set theory]]
[[Category:Basic concepts in set theory]]
[[Category:Functions and mappings]]
[[Category:Isomorphism theorems]]
[[Category:Mathematical identities]]
[[Category:Mathematics-related lists]]
[[Category:Mathematical relations]]
[[Category:Set theory]]
[[Category:Theorems in the foundations of mathematics]]

{{Sourceattribution|List of set identities and relations}}