Limit point

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Short description: A point x in a topological space, all of whose neighborhoods contain some other point in a given subset that is different from x

In mathematics, a limit point (or cluster point or accumulation point) of a set [math]\displaystyle{ S }[/math] in a topological space [math]\displaystyle{ X }[/math] is a point [math]\displaystyle{ x }[/math] that can be "approximated" by points of [math]\displaystyle{ S }[/math] in the sense that every neighbourhood of [math]\displaystyle{ x }[/math] with respect to the topology on [math]\displaystyle{ X }[/math] also contains a point of [math]\displaystyle{ S }[/math] other than [math]\displaystyle{ x }[/math] itself. A limit point of a set [math]\displaystyle{ S }[/math] does not itself have to be an element of [math]\displaystyle{ S. }[/math] There is also a closely related concept for sequences. A cluster point or accumulation point of a sequence [math]\displaystyle{ (x_n)_{n \in \mathbb{N}} }[/math] in a topological space [math]\displaystyle{ X }[/math] is a point [math]\displaystyle{ x }[/math] such that, for every neighbourhood [math]\displaystyle{ V }[/math] of [math]\displaystyle{ x, }[/math] there are infinitely many natural numbers [math]\displaystyle{ n }[/math] such that [math]\displaystyle{ x_n \in V. }[/math] This definition of a cluster or accumulation point of a sequence generalizes to nets and filters. In contrast to sets, for a sequence, net, or filter, the term "limit point" is not synonymous with a "cluster/accumulation point"; by definition, the similarly named notion of a limit point of a filter[1] (respectively, a limit point of a sequence,[2] a limit point of a net) refers to a point that the filter converges to (respectively, the sequence converges to, the net converges to).

The limit points of a set should not be confused with adherent points for which every neighbourhood of [math]\displaystyle{ x }[/math] contains a point of [math]\displaystyle{ S }[/math]. Unlike for limit points, this point of [math]\displaystyle{ S }[/math] may be [math]\displaystyle{ x }[/math] itself. A limit point can be characterized as an adherent point that is not an isolated point.

Limit points of a set should also not be confused with boundary points. For example, [math]\displaystyle{ 0 }[/math] is a boundary point (but not a limit point) of set [math]\displaystyle{ \{ 0 \} }[/math] in [math]\displaystyle{ \R }[/math] with standard topology. However, [math]\displaystyle{ 0.5 }[/math] is a limit point (though not a boundary point) of interval [math]\displaystyle{ [0, 1] }[/math] in [math]\displaystyle{ \R }[/math] with standard topology (for a less trivial example of a limit point, see the first caption).[3][4][5]

This concept profitably generalizes the notion of a limit and is the underpinning of concepts such as closed set and topological closure. Indeed, a set is closed if and only if it contains all of its limit points, and the topological closure operation can be thought of as an operation that enriches a set by uniting it with its limit points.

With respect to the usual Euclidean topology, the sequence of rational numbers [math]\displaystyle{ x_n=(-1)^n \frac{n}{n+1} }[/math] has no limit (i.e. does not converge), but has two accumulation points (which are considered limit points here), viz. -1 and +1. Thus, thinking of sets, these points are limit points of the set [math]\displaystyle{ S = \{x_n\}. }[/math]

Definition

Accumulation points of a set

Let [math]\displaystyle{ S }[/math] be a subset of a topological space [math]\displaystyle{ X. }[/math] A point [math]\displaystyle{ x }[/math] in [math]\displaystyle{ X }[/math] is a limit point or cluster point or accumulation point of the set [math]\displaystyle{ S }[/math] if every neighbourhood of [math]\displaystyle{ x }[/math] contains at least one point of [math]\displaystyle{ S }[/math] different from [math]\displaystyle{ x }[/math] itself.

It does not make a difference if we restrict the condition to open neighbourhoods only. It is often convenient to use the "open neighbourhood" form of the definition to show that a point is a limit point and to use the "general neighbourhood" form of the definition to derive facts from a known limit point.

If [math]\displaystyle{ X }[/math] is a [math]\displaystyle{ T 1 }[/math] space (such as a metric space), then [math]\displaystyle{ x \in X }[/math] is a limit point of [math]\displaystyle{ S }[/math] if and only if every neighbourhood of [math]\displaystyle{ x }[/math] contains infinitely many points of [math]\displaystyle{ S. }[/math][6] In fact, [math]\displaystyle{ T_1 }[/math] spaces are characterized by this property.

If [math]\displaystyle{ X }[/math] is a Fréchet–Urysohn space (which all metric spaces and first-countable spaces are), then [math]\displaystyle{ x \in X }[/math] is a limit point of [math]\displaystyle{ S }[/math] if and only if there is a sequence of points in [math]\displaystyle{ S \setminus \{ x \} }[/math] whose limit is [math]\displaystyle{ x. }[/math] In fact, Fréchet–Urysohn spaces are characterized by this property.

The set of limit points of [math]\displaystyle{ S }[/math] is called the derived set of [math]\displaystyle{ S. }[/math]

Types of accumulation points

If every neighbourhood of [math]\displaystyle{ x }[/math] contains infinitely many points of [math]\displaystyle{ S, }[/math] then [math]\displaystyle{ x }[/math] is a specific type of limit point called an ω-accumulation point of [math]\displaystyle{ S. }[/math]

If every neighbourhood of [math]\displaystyle{ x }[/math] contains uncountably many points of [math]\displaystyle{ S, }[/math] then [math]\displaystyle{ x }[/math] is a specific type of limit point called a condensation point of [math]\displaystyle{ S. }[/math]

If every neighbourhood [math]\displaystyle{ U }[/math] of [math]\displaystyle{ x }[/math] satisfies [math]\displaystyle{ \left| U \cap S\right| = \left| S \right|, }[/math] then [math]\displaystyle{ x }[/math] is a specific type of limit point called a complete accumulation point of [math]\displaystyle{ S. }[/math]

Accumulation points of sequences and nets

A sequence enumerating all positive rational numbers. Each positive real number is a cluster point.

In a topological space [math]\displaystyle{ X, }[/math] a point [math]\displaystyle{ x \in X }[/math] is said to be a cluster point or accumulation point of a sequence [math]\displaystyle{ x_{\bull} = \left(x_n\right)_{n=1}^{\infty} }[/math] if, for every neighbourhood [math]\displaystyle{ V }[/math] of [math]\displaystyle{ x, }[/math] there are infinitely many [math]\displaystyle{ n \in \mathbb{N} }[/math] such that [math]\displaystyle{ x_n \in V. }[/math] It is equivalent to say that for every neighbourhood [math]\displaystyle{ V }[/math] of [math]\displaystyle{ x }[/math] and every [math]\displaystyle{ n_0 \in \mathbb{N}, }[/math] there is some [math]\displaystyle{ n \geq n_0 }[/math] such that [math]\displaystyle{ x_n \in V. }[/math] If [math]\displaystyle{ X }[/math] is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then [math]\displaystyle{ x }[/math] is cluster point of [math]\displaystyle{ x_{\bull} }[/math] if and only if [math]\displaystyle{ x }[/math] is a limit of some subsequence of [math]\displaystyle{ x_{\bull}. }[/math] The set of all cluster points of a sequence is sometimes called the limit set.

Note that there is already the notion of limit of a sequence to mean a point [math]\displaystyle{ x }[/math] to which the sequence converges (that is, every neighborhood of [math]\displaystyle{ x }[/math] contains all but finitely many elements of the sequence). That is why we do not use the term limit point of a sequence as a synonym for accumulation point of the sequence.

The concept of a net generalizes the idea of a sequence. A net is a function [math]\displaystyle{ f : (P,\leq) \to X, }[/math] where [math]\displaystyle{ (P,\leq) }[/math] is a directed set and [math]\displaystyle{ X }[/math] is a topological space. A point [math]\displaystyle{ x \in X }[/math] is said to be a cluster point or accumulation point of a net [math]\displaystyle{ f }[/math] if, for every neighbourhood [math]\displaystyle{ V }[/math] of [math]\displaystyle{ x }[/math] and every [math]\displaystyle{ p_0 \in P, }[/math] there is some [math]\displaystyle{ p \geq p_0 }[/math] such that [math]\displaystyle{ f(p) \in V, }[/math] equivalently, if [math]\displaystyle{ f }[/math] has a subnet which converges to [math]\displaystyle{ x. }[/math] Cluster points in nets encompass the idea of both condensation points and ω-accumulation points. Clustering and limit points are also defined for filters.

Relation between accumulation point of a sequence and accumulation point of a set

Every sequence [math]\displaystyle{ x_{\bull} = \left(x_n\right)_{n=1}^{\infty} }[/math] in [math]\displaystyle{ X }[/math] is by definition just a map [math]\displaystyle{ x_{\bull} : \mathbb{N} \to X }[/math] so that its image [math]\displaystyle{ \operatorname{Im} x_{\bull} := \left\{ x_n : n \in \mathbb{N} \right\} }[/math] can be defined in the usual way.

  • If there exists an element [math]\displaystyle{ x \in X }[/math] that occurs infinitely many times in the sequence, [math]\displaystyle{ x }[/math] is an accumulation point of the sequence. But [math]\displaystyle{ x }[/math] need not be an accumulation point of the corresponding set [math]\displaystyle{ \operatorname{Im} x_{\bull}. }[/math] For example, if the sequence is the constant sequence with value [math]\displaystyle{ x, }[/math] we have [math]\displaystyle{ \operatorname{Im} x_{\bull} = \{ x \} }[/math] and [math]\displaystyle{ x }[/math] is an isolated point of [math]\displaystyle{ \operatorname{Im} x_{\bull} }[/math] and not an accumulation point of [math]\displaystyle{ \operatorname{Im} x_{\bull}. }[/math]
  • If no element occurs infinitely many times in the sequence, for example if all the elements are distinct, any accumulation point of the sequence is an [math]\displaystyle{ \omega }[/math]-accumulation point of the associated set [math]\displaystyle{ \operatorname{Im} x_{\bull}. }[/math]

Conversely, given a countable infinite set [math]\displaystyle{ A \subseteq X }[/math] in [math]\displaystyle{ X, }[/math] we can enumerate all the elements of [math]\displaystyle{ A }[/math] in many ways, even with repeats, and thus associate with it many sequences [math]\displaystyle{ x_{\bull} }[/math] that will satisfy [math]\displaystyle{ A = \operatorname{Im} x_{\bull}. }[/math]

  • Any [math]\displaystyle{ \omega }[/math]-accumulation point of [math]\displaystyle{ A }[/math] is an accumulation point of any of the corresponding sequences (because any neighborhood of the point will contain infinitely many elements of [math]\displaystyle{ A }[/math] and hence also infinitely many terms in any associated sequence).
  • A point [math]\displaystyle{ x \in X }[/math] that is not an [math]\displaystyle{ \omega }[/math]-accumulation point of [math]\displaystyle{ A }[/math] cannot be an accumulation point of any of the associated sequences without infinite repeats (because [math]\displaystyle{ x }[/math] has a neighborhood that contains only finitely many (possibly even none) points of [math]\displaystyle{ A }[/math] and that neighborhood can only contain finitely many terms of such sequences).

Properties

Every limit of a non-constant sequence is an accumulation point of the sequence. And by definition, every limit point is an adherent point.

The closure [math]\displaystyle{ \operatorname{cl}(S) }[/math] of a set [math]\displaystyle{ S }[/math] is a disjoint union of its limit points [math]\displaystyle{ L(S) }[/math] and isolated points [math]\displaystyle{ I(S) }[/math]:

[math]\displaystyle{ \operatorname{cl} (S) = L(S) \cup I(S), L(S) \cap I(S) = \varnothing. }[/math]

A point [math]\displaystyle{ x \in X }[/math] is a limit point of [math]\displaystyle{ S \subseteq X }[/math] if and only if it is in the closure of [math]\displaystyle{ S \setminus \{ x \}. }[/math]

Proof

We use the fact that a point is in the closure of a set if and only if every neighborhood of the point meets the set. Now, [math]\displaystyle{ x }[/math] is a limit point of [math]\displaystyle{ S, }[/math] if and only if every neighborhood of [math]\displaystyle{ x }[/math] contains a point of [math]\displaystyle{ S }[/math] other than [math]\displaystyle{ x, }[/math] if and only if every neighborhood of [math]\displaystyle{ x }[/math] contains a point of [math]\displaystyle{ S \setminus \{x\}, }[/math] if and only if [math]\displaystyle{ x }[/math] is in the closure of [math]\displaystyle{ S \setminus \{x\}. }[/math]

If we use [math]\displaystyle{ L(S) }[/math] to denote the set of limit points of [math]\displaystyle{ S, }[/math] then we have the following characterization of the closure of [math]\displaystyle{ S }[/math]: The closure of [math]\displaystyle{ S }[/math] is equal to the union of [math]\displaystyle{ S }[/math] and [math]\displaystyle{ L(S). }[/math] This fact is sometimes taken as the definition of closure.

Proof

("Left subset") Suppose [math]\displaystyle{ x }[/math] is in the closure of [math]\displaystyle{ S. }[/math] If [math]\displaystyle{ x }[/math] is in [math]\displaystyle{ S, }[/math] we are done. If [math]\displaystyle{ x }[/math] is not in [math]\displaystyle{ S, }[/math] then every neighbourhood of [math]\displaystyle{ x }[/math] contains a point of [math]\displaystyle{ S, }[/math] and this point cannot be [math]\displaystyle{ x. }[/math] In other words, [math]\displaystyle{ x }[/math] is a limit point of [math]\displaystyle{ S }[/math] and [math]\displaystyle{ x }[/math] is in [math]\displaystyle{ L(S). }[/math] ("Right subset") If [math]\displaystyle{ x }[/math] is in [math]\displaystyle{ S, }[/math] then every neighbourhood of [math]\displaystyle{ x }[/math] clearly meets [math]\displaystyle{ S, }[/math] so [math]\displaystyle{ x }[/math] is in the closure of [math]\displaystyle{ S. }[/math] If [math]\displaystyle{ x }[/math] is in [math]\displaystyle{ L(S), }[/math] then every neighbourhood of [math]\displaystyle{ x }[/math] contains a point of [math]\displaystyle{ S }[/math] (other than [math]\displaystyle{ x }[/math]), so [math]\displaystyle{ x }[/math] is again in the closure of [math]\displaystyle{ S. }[/math] This completes the proof.

A corollary of this result gives us a characterisation of closed sets: A set [math]\displaystyle{ S }[/math] is closed if and only if it contains all of its limit points.

Proof

Proof 1: [math]\displaystyle{ S }[/math] is closed if and only if [math]\displaystyle{ S }[/math] is equal to its closure if and only if [math]\displaystyle{ S=S\cup L(S) }[/math] if and only if [math]\displaystyle{ L(S) }[/math] is contained in [math]\displaystyle{ S. }[/math]

Proof 2: Let [math]\displaystyle{ S }[/math] be a closed set and [math]\displaystyle{ x }[/math] a limit point of [math]\displaystyle{ S. }[/math] If [math]\displaystyle{ x }[/math] is not in [math]\displaystyle{ S, }[/math] then the complement to [math]\displaystyle{ S }[/math] comprises an open neighbourhood of [math]\displaystyle{ x. }[/math] Since [math]\displaystyle{ x }[/math] is a limit point of [math]\displaystyle{ S, }[/math] any open neighbourhood of [math]\displaystyle{ x }[/math] should have a non-trivial intersection with [math]\displaystyle{ S. }[/math] However, a set can not have a non-trivial intersection with its complement. Conversely, assume [math]\displaystyle{ S }[/math] contains all its limit points. We shall show that the complement of [math]\displaystyle{ S }[/math] is an open set. Let [math]\displaystyle{ x }[/math] be a point in the complement of [math]\displaystyle{ S. }[/math] By assumption, [math]\displaystyle{ x }[/math] is not a limit point, and hence there exists an open neighbourhood [math]\displaystyle{ U }[/math] of [math]\displaystyle{ x }[/math] that does not intersect [math]\displaystyle{ S, }[/math] and so [math]\displaystyle{ U }[/math] lies entirely in the complement of [math]\displaystyle{ S. }[/math] Since this argument holds for arbitrary [math]\displaystyle{ x }[/math] in the complement of [math]\displaystyle{ S, }[/math] the complement of [math]\displaystyle{ S }[/math] can be expressed as a union of open neighbourhoods of the points in the complement of [math]\displaystyle{ S. }[/math] Hence the complement of [math]\displaystyle{ S }[/math] is open.

No isolated point is a limit point of any set.

Proof

If [math]\displaystyle{ x }[/math] is an isolated point, then [math]\displaystyle{ \{x\} }[/math] is a neighbourhood of [math]\displaystyle{ x }[/math] that contains no points other than [math]\displaystyle{ x. }[/math]

A space [math]\displaystyle{ X }[/math] is discrete if and only if no subset of [math]\displaystyle{ X }[/math] has a limit point.

Proof

If [math]\displaystyle{ X }[/math] is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if [math]\displaystyle{ X }[/math] is not discrete, then there is a singleton [math]\displaystyle{ \{ x \} }[/math] that is not open. Hence, every open neighbourhood of [math]\displaystyle{ \{ x \} }[/math] contains a point [math]\displaystyle{ y \neq x, }[/math] and so [math]\displaystyle{ x }[/math] is a limit point of [math]\displaystyle{ X. }[/math]

If a space [math]\displaystyle{ X }[/math] has the trivial topology and [math]\displaystyle{ S }[/math] is a subset of [math]\displaystyle{ X }[/math] with more than one element, then all elements of [math]\displaystyle{ X }[/math] are limit points of [math]\displaystyle{ S. }[/math] If [math]\displaystyle{ S }[/math] is a singleton, then every point of [math]\displaystyle{ X \setminus S }[/math] is a limit point of [math]\displaystyle{ S. }[/math]

Proof

As long as [math]\displaystyle{ S \setminus \{ x \} }[/math] is nonempty, its closure will be [math]\displaystyle{ X. }[/math] It is only empty when [math]\displaystyle{ S }[/math] is empty or [math]\displaystyle{ x }[/math] is the unique element of [math]\displaystyle{ S. }[/math]

See also

Citations

References