# Balanced set

In linear algebra and related areas of mathematics a balanced set, circled set or disk in a vector space (over a field $\mathbb {K}$ with an absolute value function $|\cdot |$) is a set $S$ such that $aS\subseteq S$ for all scalars $a$ satisfying $|a|\leq 1.$
The balanced hull or balanced envelope of a set $S$ is the smallest balanced set containing $S.$ The balanced core of a set $S$ is the largest balanced set contained in $S.$
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 $X$ be a vector space over the field $\mathbb {K}$ of real or complex numbers.
Notation
If $S$ is a set, $a$ is a scalar, and $B\subseteq \mathbb {K}$ then let $aS=\{as:s\in S\}$ and $BS=\{bs:b\in B,s\in S\}$ and for any $0\leq r\leq \infty ,$ let $B_{r}=\{a\in \mathbb {K} :|a|<r\}\qquad {\text{ and }}\qquad B_{\leq r}=\{a\in \mathbb {K} :|a|\leq r\}.$ denote, respectively, the open ball and the closed ball of radius $r$ in the scalar field $\mathbb {K}$ centered at $0$ where $B_{0}=\varnothing ,B_{\leq 0}=\{0\},$ and $B_{\infty }=B_{\leq \infty }=\mathbb {K} .$ Every balanced subset of the field $\mathbb {K}$ is of the form $B_{\leq r}$ or $B_{r}$ for some $0\leq r\leq \infty .$
Balanced set
A subset $S$ of $X$ is called a balanced set or balanced if it satisfies any of the following equivalent conditions:
* Definition: $as\in S$ for all $s\in S$ and all scalars $a$ satisfying $|a|\leq 1.$
* $aS\subseteq S$ for all scalars $a$ satisfying $|a|\leq 1.$
* $B_{\leq 1}S\subseteq S$ (where $B_{\leq 1}:=\{a\in \mathbb {K} :|a|\leq 1\}$).
* $S=B_{\leq 1}S.$
* For every $s\in S,$ $S\cap \mathbb {K} s=B_{\leq 1}(S\cap \mathbb {K} s).$
** $\mathbb {K} s=\operatorname {span} \{s\}$ is a $0$ (if $s=0$) or $1$ (if $s\neq 0$) dimensional vector subspace of $X.$
** If $R:=S\cap \mathbb {K} s$ then the above equality becomes $R=B_{\leq 1}R,$ which is exactly the previous condition for a set to be balanced. Thus, $S$ is balanced if and only if for every $s\in S,$ $S\cap \mathbb {K} s$ is a balanced set (according to any of the previous defining conditions).
* For every 1-dimensional vector subspace $Y$ of $\operatorname {span} S,$ $S\cap Y$ is a balanced set (according to any defining condition other than this one).
* For every $s\in S,$ there exists some $0\leq r\leq \infty$ such that $S\cap \mathbb {K} s=B_{r}s$ or $S\cap \mathbb {K} s=B_{\leq r}s.$
* $S$ is a balanced subset of $\operatorname {span} S$ (according to any defining condition of "balanced" other than this one).
** Thus $S$ is a balanced subset of $X$ if and only if it is balanced subset of every (equivalently, of some) vector space over the field $\mathbb {K}$ that contains $S.$ So assuming that the field $\mathbb {K}$ is clear from context, this justifies writing "$S$ is balanced" without mentioning any vector space.
If $S$ is a convex set then this list may be extended to include:
* $aS\subseteq S$ for all scalars $a$ satisfying $|a|=1.$
If $\mathbb {K} =\mathbb {R}$ then this list may be extended to include:
* $S$ is symmetric (meaning $-S=S$) and $[0,1)S\subseteq S.$

### Balanced hull

$\operatorname {bal} S~=~\bigcup _{|a|\leq 1}aS=B_{\leq 1}S$
The balanced hull of a subset $S$ of $X,$ denoted by $\operatorname {bal} S,$ is defined in any of the following equivalent ways:
* Definition: $\operatorname {bal} S$ is the smallest (with respect to $\,\subseteq \,$) balanced subset of $X$ containing $S.$
* $\operatorname {bal} S$ is the intersection of all balanced sets containing $S.$
* $\operatorname {bal} S=\bigcup _{|a|\leq 1}(aS).$
* $\operatorname {bal} S=B_{\leq 1}S.$

### Balanced core

$\operatorname {balcore} S~=~{\begin{cases}\displaystyle \bigcap _{|a|\geq 1}aS&{\text{ if }}0\in S\\\varnothing &{\text{ if }}0\not \in S\\\end{cases}}$
The balanced core of a subset $S$ of $X,$ denoted by $\operatorname {balcore} S,$ is defined in any of the following equivalent ways:
* Definition: $\operatorname {balcore} S$ is the largest (with respect to $\,\subseteq \,$) balanced subset of $S.$
* $\operatorname {balcore} S$ is the union of all balanced subsets of $S.$
* $\operatorname {balcore} S=\varnothing$ if $0\not \in S$ while $\operatorname {balcore} S=\bigcap _{|a|\geq 1}(aS)$ if $0\in S.$

## Examples

The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, $\{0\}$ 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 vector spaces
The open and closed balls centered at the origin in a normed vector space are balanced sets. If $p$ is a seminorm (or norm) on a vector space $X$ then for any constant $c>0,$ the set $\{x\in X:p(x)\leq c\}$ is balanced.
If $S\subseteq X$ is any subset and $B_{1}:=\{a\in \mathbb {K} :|a|<1\}$ then $B_{1}S$ is a balanced set. In particular, if $U\subseteq X$ is any balanced neighborhood of the origin in a topological vector space $X$ then $\operatorname {Int} _{X}U~\subseteq ~B_{1}U~=~\bigcup _{0<|a|<1}aU~\subseteq ~U.$
Balanced sets in $\mathbb {R}$ and $\mathbb {C}$
Let $\mathbb {K}$ be the field real numbers $\mathbb {R}$ or complex numbers $\mathbb {C} ,$ let $|\cdot |$ denote the absolute value on $\mathbb {K} ,$ and let $X:=\mathbb {K}$ denotes the vector space over $\mathbb {K} .$ So for example, if $\mathbb {K} :=\mathbb {C}$ is the field of complex numbers then $X=\mathbb {K} =\mathbb {C}$ is a 1-dimensional complex vector space whereas if $\mathbb {K} :=\mathbb {R}$ then $X=\mathbb {K} =\mathbb {R}$ is a 1-dimensional real vector space.
The balanced subsets of $X=\mathbb {K}$ are exactly the following:
* $\varnothing$
* $X$
* $\{0\}$
* $\{x\in X:|x|<r\}$ for some real $r>0$
* $\{x\in X:|x|\leq r\}$ for some real $r>0.$
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 $\mathbb {C}$ 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, $\mathbb {C}$ and $\mathbb {R} ^{2}$ are entirely different as far as scalar multiplication is concerned.
Balanced sets in $\mathbb {R} ^{2}$
Throughout, let $X=\mathbb {R} ^{2}$ (so $X$ is a vector space over $\mathbb {R}$) and let $B_{\leq 1}$ is the closed unit ball in $X$ centered at the origin.
If $x_{0}\in X=\mathbb {R} ^{2}$ is non-zero, and $L:=\mathbb {R} x_{0},$ then the set $R:=B_{\leq 1}\cup L$ is a closed, symmetric, and balanced neighborhood of the origin in $X.$ More generally, if $C$ is any closed subset of $X$ such that $(0,1)C\subseteq C,$ then $S:=B_{\leq 1}\cup C\cup (-C)$ is a closed, symmetric, and balanced neighborhood of the origin in $X.$ This example can be generalized to $\mathbb {R} ^{n}$ for any integer $n\geq 1.$
Let $B\subseteq \mathbb {R} ^{2}$ be the union of the line segment between the points $(-1,0)$ and $(1,0)$ and the line segment between $(0,-1)$ and $(0,1).$ Then $B$ is balanced but not convex. Nor is $B$ is absorbing (despite the fact that $\operatorname {span} B=\mathbb {R} ^{2}$ is the entire vector space).
For every $0\leq t\leq \pi ,$ let $r_{t}$ be any positive real number and let $B^{t}$ be the (open or closed) line segment in $X:=\mathbb {R} ^{2}$ between the points $(\cos t,\sin t)$ and $-(\cos t,\sin t).$ Then the set $B=\bigcup _{0\leq t<\pi }r_{t}B^{t}$ 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 $xy=1$ in $X=\mathbb {R} ^{2}.$
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 $S:=[-1,1]\times \{1\},$ which is a horizontal closed line segment lying above the $x-$axis in $X:=\mathbb {R} ^{2}.$ The balanced hull $\operatorname {bal} S$ is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles $T_{1}$ and $T_{2},$ where $T_{2}=-T_{1}$ and $T_{1}$ is the filled triangle whose vertices are the origin together with the endpoints of $S$ (said differently, $T_{1}$ is the convex hull of $S\cup \{(0,0)\}$ while $T_{2}$ is the convex hull of $(-S)\cup \{(0,0)\}$).

### Sufficient conditions

A set $T$ is balanced if and only if it is equal to its balanced hull $\operatorname {bal} T$ or to its balanced core $\operatorname {balcore} T,$ in which case all three of these sets are equal: $T=\operatorname {bal} T=\operatorname {balcore} T.$
The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field $\mathbb {K}$).
* The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.
* 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 $L:X\to Y$ is a linear map and $B\subseteq X$ and $C\subseteq Y$ are balanced sets, then $L(B)$ and $L^{-1}(C)$ are balanced sets.

### Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced. The union of the origin $\{0\}$ and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced. However, $\left\{(z,w)\in \mathbb {C} ^{2}:|z|\leq |w|\right\}$ is a balanced subset of $X=\mathbb {C} ^{2}$ that contains the origin $(0,0)\in X$ but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set. Similarly for real vector spaces, if $T$ denotes the convex hull of $(0,0)$ and $(\pm 1,1)$ (a filled triangle whose vertices are these three points) then $B:=T\cup (-T)$ is an (hour glass shaped) balanced subset of $X:=\mathbb {R} ^{2}$ whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set $\{(0,0)\}\cup \operatorname {Int} _{X}B$ 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 $X$ contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given $W\subseteq X,$ the symmetric set $\bigcap _{|u|=1}uW\subseteq W$ will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of $X$) whenever this is true of $W.$ It will be a balanced set if $W$ is a star shaped at the origin, which is true, for instance, when $W$ is convex and contains $0.$ In particular, if $W$ is a convex neighborhood of the origin then $\bigcap _{|u|=1}uW$ will be a balanced convex neighborhood of the origin and so its topological interior will be a balanced convex open neighborhood of the origin.
Let $0\in W\subseteq X$ and define $A=\bigcap _{|u|=1}uW$ (where $u$ denotes elements of the field $\mathbb {K}$ of scalars). Taking $u:=1$ shows that $A\subseteq W.$ If $W$ is convex then so is $A$ (since an intersection of convex sets is convex) and thus so is $A$'s interior. If $|s|=1$ then $sA=\bigcap _{|u|=1}suW\subseteq \bigcap _{|u|=1}uW=A$ and thus $sA=A.$ If $W$ is star shaped at the origin then so is every $uW$ (for $|u|=1$), which implies that for any $0\leq r\leq 1,$ $rA=\bigcap _{|u|=1}ruW\subseteq \bigcap _{|u|=1}uW=A$ thus proving that $A$ is balanced. If $W$ is convex and contains the origin then it is star shaped at the origin and so $A$ will be balanced.
Now suppose $W$ is a neighborhood of the origin in $X.$ Since scalar multiplication $M:\mathbb {K} \times X\to X$ (defined by $M(a,x)=ax$) is continuous at the origin $(0,0)\in \mathbb {K} \times X$ and $M(0,0)=0\in W,$ there exists some basic open neighborhood $B_{r}\times V$ (where $r>0$ and $B_{r}:=\{c\in \mathbb {K} :|c|<r\}$) of the origin in the product topology on $\mathbb {K} \times X$ such that $M\left(B_{r}\times V\right)\subseteq W;$ the set $M\left(B_{r}\times V\right)=B_{r}V$ is balanced and it is also open because it may be written as $B_{r}V=\bigcup _{|a|<r}aV=\bigcup _{0<|a|<r}aV\qquad {\text{ (since }}0\cdot V=\{0\}\subseteq aV{\text{ )}}$ where $aV$ is an open neighborhood of the origin whenever $a\neq 0.$ Finally, $A=\bigcap _{|u|=1}uW\supseteq \bigcap _{|u|=1}uB_{r}V=\bigcap _{|u|=1}B_{r}V=B_{r}V$ shows that $A$ is also a neighborhood of the origin. If $A$ is balanced then because its interior $\operatorname {Int} _{X}A$ contains the origin, $\operatorname {Int} _{X}A$ will also be balanced. If $W$ is convex then $A$ is convex and balanced and thus the same is true of $\operatorname {Int} _{X}A.$ $\blacksquare$
Suppose that $W$ is a convex and absorbing subset of $X.$ Then $D:=\bigcap _{|u|=1}uW$ will be convex balanced absorbing subset of $X,$ which guarantees that the Minkowski functional $p_{D}:X\to \mathbb {R}$ of $D$ will be a seminorm on $X,$ thereby making $\left(X,p_{D}\right)$ into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples $rD$ as $r$ ranges over $\left\{{\tfrac {1}{2}},{\tfrac {1}{3}},{\tfrac {1}{4}},\ldots \right\}$ (or over any other set of non-zero scalars having $0$ as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If $X$ is a topological vector space and if this convex absorbing subset $W$ is also a bounded subset of $X,$ then the same will be true of the absorbing disk $D:={\textstyle \bigcap \limits _{|u|=1}}uW;$ if in addition $D$ does not contain any non-trivial vector subspace then $p_{D}$ will be a norm and $\left(X,p_{D}\right)$ will form what is known as an auxiliary normed space. If this normed space is a Banach space then $D$ 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 $B$ is a balanced subset of $X$ then:
* for any scalars $c$ and $d,$ if $|c|\leq |d|$ then $cB\subseteq dB$ and $cB=|c|B.$ Thus if $c$ and $d$ are any scalars then $(cB)\cap (dB)=\min _\{|c|,|d|\}B.$
* $B$ is absorbing in $X$ if and only if for all $x\in X,$ there exists $r>0$ such that $x\in rB.$
* for any 1-dimensional vector subspace $Y$ of $X,$ the set $B\cap Y$ is convex and balanced. If $B$ is not empty and if $Y$ is a 1-dimensional vector subspace of $\operatorname {span} B$ then $B\cap Y$ is either $\{0\}$ or else it is absorbing in $Y.$
* for any $x\in X,$ if $B\cap \operatorname {span} x$ contains more than one point then it is a convex and balanced neighborhood of $0$ in the 1-dimensional vector space $\operatorname {span} x$ when this space is endowed with the Hausdorff Euclidean topology; and the set $B\cap \mathbb {R} x$ is a convex balanced subset of the real vector space $\mathbb {R} x$ that contains the origin.
Properties of balanced hulls and balanced cores
For any collection ${\mathcal {S}}$ of subsets of $X,$ $\operatorname {bal} \left(\bigcup _{S\in {\mathcal {S}}}S\right)=\bigcup _{S\in {\mathcal {S}}}\operatorname {bal} S\quad {\text{ and }}\quad \operatorname {balcore} \left(\bigcap _{S\in {\mathcal {S}}}S\right)=\bigcap _{S\in {\mathcal {S}}}\operatorname {balcore} S.$
In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If $X$ is a Hausdorff topological vector space and if $K$ is a compact subset of $X$ then the balanced hull of $K$ is compact.
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 $S\subseteq X$ and any scalar $c,$ $\operatorname {bal} (c\,S)=c\operatorname {bal} S=|c|\operatorname {bal} S.$
For any scalar $c\neq 0,$ $\operatorname {balcore} (c\,S)=c\operatorname {balcore} S=|c|\operatorname {balcore} S.$ This equality holds for $c=0$ if and only if $S\subseteq \{0\}.$ Thus if $0\in S$ or $S=\varnothing$ then $\operatorname {balcore} (c\,S)=c\operatorname {balcore} S=|c|\operatorname {balcore} S$ for every scalar $c.$

## Related notions

A function $p:X\to [0,\infty )$ on a real or complex vector space is said to be a balanced function if it satisfies any of the following equivalent conditions:
* $p(ax)\leq p(x)$ whenever $a$ is a scalar satisfying $|a|\leq 1$ and $x\in X.$
* $p(ax)\leq p(bx)$ whenever $a$ and $b$ are scalars satisfying $|a|\leq |b|$ and $x\in X.$
* $\{x\in X:p(x)\leq t\}$ is a balanced set for every non-negative real $t\geq 0.$
If $p$ is a balanced function then $p(ax)=p(|a|x)$ for every scalar $a$ and vector $x\in X;$ so in particular, $p(ux)=p(x)$ for every unit length scalar $u$ (satisfying $|u|=1$) and every $x\in X.$ Using $u:=-1$ shows that every balanced function is a symmetric function.
A real-valued function $p:X\to \mathbb {R}$ is a seminorm if and only if it is a balanced sublinear function.