# Balanced set

Short description: Construct in functional analysis

In linear algebra and related areas of mathematics a balanced set, circled set or disk in a vector space (over a field $\displaystyle{ \mathbb{K} }$ with an absolute value function $\displaystyle{ |\cdot | }$) is a set $\displaystyle{ S }$ such that $\displaystyle{ a S \subseteq S }$ for all scalars $\displaystyle{ a }$ satisfying $\displaystyle{ |a| \leq 1. }$

The balanced hull or balanced envelope of a set $\displaystyle{ S }$ is the smallest balanced set containing $\displaystyle{ S. }$ The balanced core of a set $\displaystyle{ S }$ is the largest balanced set contained in $\displaystyle{ S. }$

Balanced sets are ubiquitous in functional analysis because every neighborhood of the origin in every topological vector space (TVS) contains a balanced neighborhood of the origin and every convex neighborhood of the origin contains a balanced convex neighborhood of the origin (even if the TVS is not locally convex). This neighborhood can also be chosen to be an open set or, alternatively, a closed set.

## Definition

Let $\displaystyle{ X }$ be a vector space over the field $\displaystyle{ \mathbb{K} }$ of real or complex numbers.

Notation

If $\displaystyle{ S }$ is a set, $\displaystyle{ a }$ is a scalar, and $\displaystyle{ B \subseteq \mathbb{K} }$ then let $\displaystyle{ a S = \{a s : s \in S\} }$ and $\displaystyle{ B S = \{b s : b \in B, s \in S\} }$ and for any $\displaystyle{ 0 \leq r \leq \infty, }$ let $\displaystyle{ B_r = \{a \in \mathbb{K} : |a| \lt r\} \qquad \text{ and } \qquad B_{\leq r} = \{ a \in \mathbb{K} : |a| \leq r\}. }$ denote, respectively, the open ball and the closed ball of radius $\displaystyle{ r }$ in the scalar field $\displaystyle{ \mathbb{K} }$ centered at $\displaystyle{ 0 }$ where $\displaystyle{ B_0 = \varnothing, B_{\leq 0} = \{0\}, }$ and $\displaystyle{ B_{\infty} = B_{\leq \infty} = \mathbb{K}. }$ Every balanced subset of the field $\displaystyle{ \mathbb{K} }$ is of the form $\displaystyle{ B_{\leq r} }$ or $\displaystyle{ B_r }$ for some $\displaystyle{ 0 \leq r \leq \infty. }$

Balanced set

A subset $\displaystyle{ S }$ of $\displaystyle{ X }$ is called a balanced set or balanced if it satisfies any of the following equivalent conditions:

1. Definition: $\displaystyle{ a s \in S }$ for all $\displaystyle{ s \in S }$ and all scalars $\displaystyle{ a }$ satisfying $\displaystyle{ |a| \leq 1. }$
2. $\displaystyle{ a S \subseteq S }$ for all scalars $\displaystyle{ a }$ satisfying $\displaystyle{ |a| \leq 1. }$
3. $\displaystyle{ B_{\leq 1} S \subseteq S }$ (where $\displaystyle{ B_{\leq 1} := \{a \in \mathbb{K} : |a| \leq 1\} }$).
4. $\displaystyle{ S = B_{\leq 1} S. }$[1]
5. For every $\displaystyle{ s \in S, }$ $\displaystyle{ S \cap \mathbb{K} s = B_{\leq 1} (S \cap \mathbb{K} s). }$
• $\displaystyle{ \mathbb{K} s = \operatorname{span} \{s\} }$ is a $\displaystyle{ 0 }$ (if $\displaystyle{ s = 0 }$) or $\displaystyle{ 1 }$ (if $\displaystyle{ s \neq 0 }$) dimensional vector subspace of $\displaystyle{ X. }$
• If $\displaystyle{ R := S \cap \mathbb{K} s }$ then the above equality becomes $\displaystyle{ R = B_{\leq 1} R, }$ which is exactly the previous condition for a set to be balanced. Thus, $\displaystyle{ S }$ is balanced if and only if for every $\displaystyle{ s \in S, }$ $\displaystyle{ S \cap \mathbb{K} s }$ is a balanced set (according to any of the previous defining conditions).
6. For every 1-dimensional vector subspace $\displaystyle{ Y }$ of $\displaystyle{ \operatorname{span} S, }$ $\displaystyle{ S \cap Y }$ is a balanced set (according to any defining condition other than this one).
7. For every $\displaystyle{ s \in S, }$ there exists some $\displaystyle{ 0 \leq r \leq \infty }$ such that $\displaystyle{ S \cap \mathbb{K} s = B_r s }$ or $\displaystyle{ S \cap \mathbb{K} s = B_{\leq r} s. }$
8. $\displaystyle{ S }$ is a balanced subset of $\displaystyle{ \operatorname{span} S }$ (according to any defining condition of "balanced" other than this one).
• Thus $\displaystyle{ S }$ is a balanced subset of $\displaystyle{ X }$ if and only if it is balanced subset of every (equivalently, of some) vector space over the field $\displaystyle{ \mathbb{K} }$ that contains $\displaystyle{ S. }$ So assuming that the field $\displaystyle{ \mathbb{K} }$ is clear from context, this justifies writing "$\displaystyle{ S }$ is balanced" without mentioning any vector space.[note 1]

If $\displaystyle{ S }$ is a convex set then this list may be extended to include:

1. $\displaystyle{ a S \subseteq S }$ for all scalars $\displaystyle{ a }$ satisfying $\displaystyle{ |a| = 1. }$[2]

If $\displaystyle{ \mathbb{K} = \R }$ then this list may be extended to include:

1. $\displaystyle{ S }$ is symmetric (meaning $\displaystyle{ - S = S }$) and $\displaystyle{ [0, 1) S \subseteq S. }$

### Balanced hull

$\displaystyle{ \operatorname{bal} S ~=~ \bigcup_{|a| \leq 1} a S = B_{\leq 1} S }$

The balanced hull of a subset $\displaystyle{ S }$ of $\displaystyle{ X, }$ denoted by $\displaystyle{ \operatorname{bal} S, }$ is defined in any of the following equivalent ways:

1. Definition: $\displaystyle{ \operatorname{bal} S }$ is the smallest (with respect to $\displaystyle{ \,\subseteq\, }$) balanced subset of $\displaystyle{ X }$ containing $\displaystyle{ S. }$
2. $\displaystyle{ \operatorname{bal} S }$ is the intersection of all balanced sets containing $\displaystyle{ S. }$
3. $\displaystyle{ \operatorname{bal} S = \bigcup_{|a| \leq 1} (a S). }$
4. $\displaystyle{ \operatorname{bal} S = B_{\leq 1} S. }$[1]

### Balanced core

$\displaystyle{ \operatorname{balcore} S ~=~ \begin{cases} \displaystyle\bigcap_{|a| \geq 1} a S & \text{ if } 0 \in S \\ \varnothing & \text{ if } 0 \not\in S \\ \end{cases} }$

The balanced core of a subset $\displaystyle{ S }$ of $\displaystyle{ X, }$ denoted by $\displaystyle{ \operatorname{balcore} S, }$ is defined in any of the following equivalent ways:

1. Definition: $\displaystyle{ \operatorname{balcore} S }$ is the largest (with respect to $\displaystyle{ \,\subseteq\, }$) balanced subset of $\displaystyle{ S. }$
2. $\displaystyle{ \operatorname{balcore} S }$ is the union of all balanced subsets of $\displaystyle{ S. }$
3. $\displaystyle{ \operatorname{balcore} S = \varnothing }$ if $\displaystyle{ 0 \not\in S }$ while $\displaystyle{ \operatorname{balcore} S = \bigcap_{|a| \geq 1} (a S) }$ if $\displaystyle{ 0 \in S. }$

## Examples

The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, $\displaystyle{ \{0\} }$ is always a balanced set.

Any non-empty set that does not contain the origin is not balanced and furthermore, the balanced core of such a set will equal the empty set.

Normed and topological vectors spaces

The open and closed balls centered at the origin in a normed vector space are balanced sets. If $\displaystyle{ p }$ is a seminorm (or norm) on a vector space $\displaystyle{ X }$ then for any constant $\displaystyle{ c \gt 0, }$ the set $\displaystyle{ \{x \in X : p(x) \leq c\} }$ is balanced.

If $\displaystyle{ S \subseteq X }$ is any subset and $\displaystyle{ B_1 := \{a \in \mathbb{K} : |a| \lt 1\} }$ then $\displaystyle{ B_1 S }$ is a balanced set. In particular, if $\displaystyle{ U \subseteq X }$ is any balanced neighborhood of the origin in a topological vector space $\displaystyle{ X }$ then $\displaystyle{ \operatorname{Int}_X U ~\subseteq~ B_1 U ~=~ \bigcup_{0 \lt |a| \lt 1} a U ~\subseteq~ U. }$

Balanced sets in $\displaystyle{ \R }$ and $\displaystyle{ \Complex }$

Let $\displaystyle{ \mathbb{K} }$ be the field real numbers $\displaystyle{ \R }$ or complex numbers $\displaystyle{ \Complex, }$ let $\displaystyle{ |\cdot| }$ denote the absolute value on $\displaystyle{ \mathbb{K}, }$ and let $\displaystyle{ X := \mathbb{K} }$ denotes the vector space over $\displaystyle{ \mathbb{K}. }$ So for example, if $\displaystyle{ \mathbb{K} := \Complex }$ is the field of complex numbers then $\displaystyle{ X = \mathbb{K} = \Complex }$ is a 1-dimensional complex vector space whereas if $\displaystyle{ \mathbb{K} := \R }$ then $\displaystyle{ X = \mathbb{K} = \R }$ is a 1-dimensional real vector space.

The balanced subsets of $\displaystyle{ X = \mathbb{K} }$ are exactly the following:[3]

1. $\displaystyle{ \varnothing }$
2. $\displaystyle{ X }$
3. $\displaystyle{ \{0\} }$
4. $\displaystyle{ \{x \in X : |x| \lt r\} }$ for some real $\displaystyle{ r \gt 0 }$
5. $\displaystyle{ \{x \in X : |x| \leq r\} }$ for some real $\displaystyle{ r \gt 0. }$

Consequently, both the balanced core and the balanced hull of every set of scalars is equal to one of the sets listed above.

The balanced sets are $\displaystyle{ \Complex }$ itself, the empty set and the open and closed discs centered at zero. Contrariwise, in the two dimensional Euclidean space there are many more balanced sets: any line segment with midpoint at the origin will do. As a result, $\displaystyle{ \Complex }$ and $\displaystyle{ \R^2 }$ are entirely different as far as scalar multiplication is concerned.

Balanced sets in $\displaystyle{ \R^2 }$

Throughout, let $\displaystyle{ X = \R^2 }$ (so $\displaystyle{ X }$ is a vector space over $\displaystyle{ \R }$) and let $\displaystyle{ B_{\leq 1} }$ is the closed unit ball in $\displaystyle{ X }$ centered at the origin.

If $\displaystyle{ x_0 \in X = \R^2 }$ is non-zero, and $\displaystyle{ L := \R x_0, }$ then the set $\displaystyle{ R := B_{\leq 1} \cup L }$ is a closed, symmetric, and balanced neighborhood of the origin in $\displaystyle{ X. }$ More generally, if $\displaystyle{ C }$ is any closed subset of $\displaystyle{ X }$ such that $\displaystyle{ (0, 1) C \subseteq C, }$ then $\displaystyle{ S := B_{\leq 1} \cup C \cup (-C) }$ is a closed, symmetric, and balanced neighborhood of the origin in $\displaystyle{ X. }$ This example can be generalized to $\displaystyle{ \R^n }$ for any integer $\displaystyle{ n \geq 1. }$

Let $\displaystyle{ B \subseteq \R^2 }$ be the union of the line segment between the points $\displaystyle{ (-1, 0) }$ and $\displaystyle{ (1, 0) }$ and the line segment between $\displaystyle{ (0, -1) }$ and $\displaystyle{ (0, 1). }$ Then $\displaystyle{ B }$ is balanced but not convex. Nor is $\displaystyle{ B }$ is absorbing (despite the fact that $\displaystyle{ \operatorname{span} B = \R^2 }$ is the entire vector space).

For every $\displaystyle{ 0 \leq t \leq \pi, }$ let $\displaystyle{ r_t }$ be any positive real number and let $\displaystyle{ B^t }$ be the (open or closed) line segment in $\displaystyle{ X := \R^2 }$ between the points $\displaystyle{ (\cos t, \sin t) }$ and $\displaystyle{ - (\cos t, \sin t). }$ Then the set $\displaystyle{ B = \bigcup_{0 \leq t \lt \pi} r_t B^t }$ is a balanced and absorbing set but it is not necessarily convex.

The balanced hull of a closed set need not be closed. Take for instance the graph of $\displaystyle{ x y = 1 }$ in $\displaystyle{ X = \R^2. }$

The next example shows that the balanced hull of a convex set may fail to be convex (however, the convex hull of a balanced set is always balanced). For an example, let the convex subset be $\displaystyle{ S := [-1, 1] \times \{1\}, }$ which is a horizontal closed line segment lying above the $\displaystyle{ x- }$axis in $\displaystyle{ X := \R^2. }$ The balanced hull $\displaystyle{ \operatorname{bal} S }$ is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles $\displaystyle{ T_1 }$ and $\displaystyle{ T_2, }$ where $\displaystyle{ T_2 = - T_1 }$ and $\displaystyle{ T_1 }$ is the filled triangle whose vertices are the origin together with the endpoints of $\displaystyle{ S }$ (said differently, $\displaystyle{ T_1 }$ is the convex hull of $\displaystyle{ S \cup \{(0,0)\} }$ while $\displaystyle{ T_2 }$ is the convex hull of $\displaystyle{ (-S) \cup \{(0,0)\} }$).

### Sufficient conditions

A set $\displaystyle{ T }$ is balanced if and only if it is equal to its balanced hull $\displaystyle{ \operatorname{bal} T }$ or to its balanced core $\displaystyle{ \operatorname{balcore} T, }$ in which case all three of these sets are equal: $\displaystyle{ T = \operatorname{bal} T = \operatorname{balcore} T. }$

The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field $\displaystyle{ \mathbb{K} }$).

• The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.[4]
• The convex hull of a balanced set is convex and balanced (that is, it is absolutely convex). However, the balanced hull of a convex set may fail to be convex (a counter-example is given above).
• Arbitrary unions of balanced sets are balanced, and the same is true of arbitrary intersections of balanced sets.
• Scalar multiples and (finite) Minkowski sums of balanced sets are again balanced.
• Images and preimages of balanced sets under linear maps are again balanced. Explicitly, if $\displaystyle{ L : X \to Y }$ is a linear map and $\displaystyle{ B \subseteq X }$ and $\displaystyle{ C \subseteq Y }$ are balanced sets, then $\displaystyle{ L(B) }$ and $\displaystyle{ L^{-1}(C) }$ are balanced sets.

### Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced.[5] The union of the origin $\displaystyle{ \{0\} }$ and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced.[5][proof 1] However, $\displaystyle{ \left\{(z, w) \in \Complex^2 : |z| \leq |w|\right\} }$ is a balanced subset of $\displaystyle{ X = \Complex^2 }$ that contains the origin $\displaystyle{ (0, 0) \in X }$ but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set.[6] Similarly for real vector spaces, if $\displaystyle{ T }$ denotes the convex hull of $\displaystyle{ (0, 0) }$ and $\displaystyle{ (\pm 1, 1) }$ (a filled triangle whose vertices are these three points) then $\displaystyle{ B := T \cup (-T) }$ is an (hour glass shaped) balanced subset of $\displaystyle{ X := \Reals^2 }$ whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set $\displaystyle{ \{(0, 0)\} \cup \operatorname{Int}_X B }$ formed by adding the origin is balanced, it is neither an open set nor a neighborhood of the origin).

Every neighborhood (respectively, convex neighborhood) of the origin in a topological vector space $\displaystyle{ X }$ contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given $\displaystyle{ W \subseteq X, }$ the symmetric set $\displaystyle{ \bigcap_{|u|=1} u W \subseteq W }$ will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of $\displaystyle{ X }$) whenever this is true of $\displaystyle{ W. }$ It will be a balanced set if $\displaystyle{ W }$ is a star shaped at the origin,[note 2] which is true, for instance, when $\displaystyle{ W }$ is convex and contains $\displaystyle{ 0. }$ In particular, if $\displaystyle{ W }$ is a convex neighborhood of the origin then $\displaystyle{ \bigcap_{|u|=1} u W }$ will be a balanced convex neighborhood of the origin and so its topological interior will be a balanced convex open neighborhood of the origin.[5]

Suppose that $\displaystyle{ W }$ is a convex and absorbing subset of $\displaystyle{ X. }$ Then $\displaystyle{ D := \bigcap_{|u|=1} u W }$ will be convex balanced absorbing subset of $\displaystyle{ X, }$ which guarantees that the Minkowski functional $\displaystyle{ p_D : X \to \R }$ of $\displaystyle{ D }$ will be a seminorm on $\displaystyle{ X, }$ thereby making $\displaystyle{ \left(X, p_D\right) }$ into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples $\displaystyle{ r D }$ as $\displaystyle{ r }$ ranges over $\displaystyle{ \left\{\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \ldots\right\} }$ (or over any other set of non-zero scalars having $\displaystyle{ 0 }$ as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If $\displaystyle{ X }$ is a topological vector space and if this convex absorbing subset $\displaystyle{ W }$ is also a bounded subset of $\displaystyle{ X, }$ then the same will be true of the absorbing disk $\displaystyle{ D := {\textstyle\bigcap\limits_{|u|=1}} u W; }$ if in addition $\displaystyle{ D }$ does not contain any non-trivial vector subspace then $\displaystyle{ p_D }$ will be a norm and $\displaystyle{ \left(X, p_D\right) }$ will form what is known as an auxiliary normed space.[7] If this normed space is a Banach space then $\displaystyle{ D }$ is called a Banach disk.

## Properties

Properties of balanced sets

A balanced set is not empty if and only if it contains the origin. By definition, a set is absolutely convex if and only if it is convex and balanced. Every balanced set is star-shaped (at 0) and a symmetric set. If $\displaystyle{ B }$ is a balanced subset of $\displaystyle{ X }$ then:

• for any scalars $\displaystyle{ c }$ and $\displaystyle{ d, }$ if $\displaystyle{ |c| \leq |d| }$ then $\displaystyle{ c B \subseteq d B }$ and $\displaystyle{ c B = |c| B. }$ Thus if $\displaystyle{ c }$ and $\displaystyle{ d }$ are any scalars then $\displaystyle{ (c B) \cap (d B) = \min_{} \{|c|, |d|\} B. }$
• $\displaystyle{ B }$ is absorbing in $\displaystyle{ X }$ if and only if for all $\displaystyle{ x \in X, }$ there exists $\displaystyle{ r \gt 0 }$ such that $\displaystyle{ x \in r B. }$[2]
• for any 1-dimensional vector subspace $\displaystyle{ Y }$ of $\displaystyle{ X, }$ the set $\displaystyle{ B \cap Y }$ is convex and balanced. If $\displaystyle{ B }$ is not empty and if $\displaystyle{ Y }$ is a 1-dimensional vector subspace of $\displaystyle{ \operatorname{span} B }$ then $\displaystyle{ B \cap Y }$ is either $\displaystyle{ \{0\} }$ or else it is absorbing in $\displaystyle{ Y. }$
• for any $\displaystyle{ x \in X, }$ if $\displaystyle{ B \cap \operatorname{span} x }$ contains more than one point then it is a convex and balanced neighborhood of $\displaystyle{ 0 }$ in the 1-dimensional vector space $\displaystyle{ \operatorname{span} x }$ when this space is endowed with the Hausdorff Euclidean topology; and the set $\displaystyle{ B \cap \R x }$ is a convex balanced subset of the real vector space $\displaystyle{ \R x }$ that contains the origin.

Properties of balanced hulls and balanced cores

For any collection $\displaystyle{ \mathcal{S} }$ of subsets of $\displaystyle{ X, }$ $\displaystyle{ \operatorname{bal} \left(\bigcup_{S \in \mathcal{S}} S\right) = \bigcup_{S \in \mathcal{S}} \operatorname{bal} S \quad \text{ and } \quad \operatorname{balcore} \left(\bigcap_{S \in \mathcal{S}} S\right) = \bigcap_{S \in \mathcal{S}} \operatorname{balcore} S. }$

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If $\displaystyle{ X }$ is a Hausdorff topological vector space and if $\displaystyle{ K }$ is a compact subset of $\displaystyle{ X }$ then the balanced hull of $\displaystyle{ K }$ is compact.[8]

If a set is closed (respectively, convex, absorbing, a neighborhood of the origin) then the same is true of its balanced core.

For any subset $\displaystyle{ S \subseteq X }$ and any scalar $\displaystyle{ c, }$ $\displaystyle{ \operatorname{bal} (c \, S) = c \operatorname{bal} S = |c| \operatorname{bal} S. }$

For any scalar $\displaystyle{ c \neq 0, }$ $\displaystyle{ \operatorname{balcore} (c \, S) = c \operatorname{balcore} S = |c| \operatorname{balcore} S. }$ This equality holds for $\displaystyle{ c = 0 }$ if and only if $\displaystyle{ S \subseteq \{0\}. }$ Thus if $\displaystyle{ 0 \in S }$ or $\displaystyle{ S = \varnothing }$ then $\displaystyle{ \operatorname{balcore} (c \, S) = c \operatorname{balcore} S = |c| \operatorname{balcore} S }$ for every scalar $\displaystyle{ c. }$

## Related notions

A function $\displaystyle{ p : X \to [0, \infty) }$ on a real or complex vector space is said to be a balanced function if it satisfies any of the following equivalent conditions:[9]

1. $\displaystyle{ p(a x) \leq p(x) }$ whenever $\displaystyle{ a }$ is a scalar satisfying $\displaystyle{ |a| \leq 1 }$ and $\displaystyle{ x \in X. }$
2. $\displaystyle{ p(a x) \leq p(b x) }$ whenever $\displaystyle{ a }$ and $\displaystyle{ b }$ are scalars satisfying $\displaystyle{ |a| \leq |b| }$ and $\displaystyle{ x \in X. }$
3. $\displaystyle{ \{x \in X : p(x) \leq t\} }$ is a balanced set for every non-negative real $\displaystyle{ t \geq 0. }$

If $\displaystyle{ p }$ is a balanced function then $\displaystyle{ p(a x) = p(|a| x) }$ for every scalar $\displaystyle{ a }$ and vector $\displaystyle{ x \in X; }$ so in particular, $\displaystyle{ p(u x) = p(x) }$ for every unit length scalar $\displaystyle{ u }$ (satisfying $\displaystyle{ |u| = 1 }$) and every $\displaystyle{ x \in X. }$[9] Using $\displaystyle{ u := -1 }$ shows that every balanced function is a symmetric function.

A real-valued function $\displaystyle{ p : X \to \R }$ is a seminorm if and only if it is a balanced sublinear function.

## References

1. Swartz 1992, pp. 4-8.
2. Narici & Beckenstein 2011, pp. 107-110.
3. Jarchow 1981, p. 34.
4. Narici & Beckenstein 2011, pp. 156-175.
5. Rudin 1991, pp. 10-14.
6. Rudin 1991, p. 38.
7. Narici & Beckenstein 2011, pp. 115-154.
8. Trèves 2006, p. 56.
9. Schechter 1996, p. 313.
1. Assuming that all vector spaces containing a set $\displaystyle{ S }$ are over the same field, when describing the set as being "balanced", it is not necessary to mention a vector space containing $\displaystyle{ S. }$ That is, "$\displaystyle{ S }$ is balanced" may be written in place of "$\displaystyle{ S }$ is a balanced subset of $\displaystyle{ X }$".
2. $\displaystyle{ W }$ being star shaped at the origin means that $\displaystyle{ 0 \in W }$ and $\displaystyle{ r w \in W }$ for all $\displaystyle{ 0 \leq r \leq 1 }$ and $\displaystyle{ w \in W. }$

Proofs

1. Let $\displaystyle{ B \subseteq X }$ be balanced. If its topological interior $\displaystyle{ \operatorname{Int}_X B }$ is empty then it is balanced so assume otherwise and let $\displaystyle{ |s| \leq 1 }$ be a scalar. If $\displaystyle{ s \neq 0 }$ then the map $\displaystyle{ X \to X }$ defined by $\displaystyle{ x \mapsto s x }$ is a homeomorphism, which implies $\displaystyle{ s \operatorname{Int}_X B = \operatorname{Int}_X (s B) \subseteq s B \subseteq B; }$ because $\displaystyle{ s \operatorname{Int}_X B }$ is open, $\displaystyle{ s \operatorname{Int}_X B \subseteq \operatorname{Int}_X B }$ so that it only remains to show that this is true for $\displaystyle{ s = 0. }$ However, $\displaystyle{ 0 \in \operatorname{Int}_X B }$ might not be true but when it is true then $\displaystyle{ \operatorname{Int}_X B }$ will be balanced. $\displaystyle{ \blacksquare }$