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 $|a|\le 1.$ 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 $0\le r\le \mathrm{\infty },$ let $$B_r=\{a\in \mathbb{𝕂}:|a|<r\}\text{ and }{B}_{\le r}=\{a\in \mathbb{𝕂}:|a|\le r\}.$$ denote, respectively, the open ball and the closed ball of radius $r$ in the scalar field $\mathbb{𝕂}$ centered at $0$ where $B_0=\varnothing ,B_{\le 0}=\{0\},$ and $B_{\mathrm{\infty }}=B_{\le \mathrm{\infty }}=\mathbb{𝕂}.$ Every balanced subset of the field $\mathbb{𝕂}$ is of the form $B_{\le r}$ or $B_r$ for some $0\le r\le \mathrm{\infty }.$ Balanced set A subset $S$ of $X$ is called a balanced set or balanced if it satisfies any of the following equivalent conditions:

1. Definition: $as\in S$ for all $s\in S$ and all scalars $a$ satisfying $|a|\le 1.$
2. $aS\subseteq S$ for all scalars $a$ satisfying $|a|\le 1.$
3. $B_{\le 1}S\subseteq S$ (where $B_{\le 1}:=\{a\in \mathbb{𝕂}:|a|\le 1\}$).
4. $S=B_{\le 1}S.$
5. For every $s\in S,$ $S\cap \mathbb{𝕂}s=B_{\le 1}(S\cap \mathbb{𝕂}s).$
   • $\mathbb{𝕂}s=\mathrm{span}\{s\}$ is a $0$ (if $s=0$) or $1$ (if $s\ne 0$) dimensional vector subspace of $X.$
   • If $R:=S\cap \mathbb{𝕂}s$ then the above equality becomes $R=B_{\le 1}R,$ 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).
6. For every 1-dimensional vector subspace $Y$ of $\mathrm{span}S,$ $S\cap Y$ is a balanced set (according to any defining condition other than this one).
7. For every $s\in S,$ there exists some $0\le r\le \mathrm{\infty }$ such that $S\cap \mathbb{𝕂}s=B_{r}s$ or $S\cap \mathbb{𝕂}s=B_{\le r}s.$
8. $S$ is a balanced subset of $\mathrm{span}S$ (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:

9. $aS\subseteq S$ for all scalars $a$ satisfying $|a|=1.$

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

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

Balanced hull

$$\mathrm{bal}S=\bigcup _{|a|\le 1}aS=B_{\le 1}S$$ The balanced hull of a subset $S$ of $X,$ denoted by $\mathrm{bal}S,$ is defined in any of the following equivalent ways:

1. Definition: $\mathrm{bal}S$ is the smallest (with respect to $\subseteq$) balanced subset of $X$ containing $S.$
2. $\mathrm{bal}S$ is the intersection of all balanced sets containing $S.$
3. $\mathrm{bal}S=\bigcup _{|a|\le 1}(aS).$
4. $\mathrm{bal}S=B_{\le 1}S.$

Balanced core

$$\mathrm{balcore}S=\{\begin{array}{ll}\bigcap _{|a|\ge 1}aS& \text{ if }0\in S\\ \varnothing & \text{ if }0\in ̸S\\ \end{array}$$ The balanced core of a subset $S$ of $X,$ denoted by $\mathrm{balcore}S,$ is defined in any of the following equivalent ways:

1. Definition: $\mathrm{balcore}S$ is the largest (with respect to $\subseteq$) balanced subset of $S.$
2. $\mathrm{balcore}S$ is the union of all balanced subsets of $S.$
3. $\mathrm{balcore}S=\varnothing$ if $0\in ̸S$ while $\mathrm{balcore}S=\bigcap _{|a|\ge 1}(aS)$ 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 $\{x\in X:p(x)\le c\}$ 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 $${\mathrm{Int}}_{X}U\subseteq B_{1}U=\bigcup _{0<|a|<1}aU\subseteq U.$$ 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:

1. $\varnothing$
2. $X$
3. $\{0\}$
4. $\{x\in X:|x|<r\}$ for some real $r>0$
5. $\{x\in X:|x|\le r\}$ 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 $B_{\le 1}$ 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 $R:=B_{\le 1}\cup L$ 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 $S:=B_{\le 1}\cup C\cup (-C)$ is a closed, symmetric, and balanced neighborhood of the origin in $X.$ This example can be generalized to ${\mathbb{ℝ}}^n$ for any integer $n\ge 1.$ 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 $\mathrm{span}B={\mathbb{ℝ}}^2$ is the entire vector space). For every $0\le t\le \pi ,$ 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 $B=\bigcup _{0\le t<\pi }{r}_t{B}^t$ 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 $\mathrm{bal}S$ 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 $\mathrm{bal}T$ or to its balanced core $\mathrm{balcore}T,$ in which case all three of these sets are equal: $T=\mathrm{bal}T=\mathrm{balcore}T.$ 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, $\{(z,w)\in {\mathbb{ℂ}}^2:|z|\le |w|\}$ 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 $\{(0,0)\}\cup {\mathrm{Int}}_{X}B$ 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.

Proof

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 $0\le r\le 1,$ $$rA=\bigcap _{|u|=1}ruW\subseteq \bigcap _{|u|=1}uW=A$$ 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 $M(B_r\times V)\subseteq W;$ the set $M(B_r\times V)=B_{r}V$ is balanced and it is also open because it may be written as $$B_{r}V=\bigcup _{|a|<r}aV=\bigcup _{0<|a|<r}aV\text{ (since }0\cdot V=\{0\}\subseteq aV\text{ )}$$ where $aV$ is an open neighborhood of the origin whenever $a\ne 0.$ Finally, $$A=\bigcap _{|u|=1}uW\supseteq \bigcap _{|u|=1}u{B}_{r}V=\bigcap _{|u|=1}{B}_{r}V=B_{r}V$$ shows that $A$ is also a neighborhood of the origin. If $A$ is balanced then because its interior ${\mathrm{Int}}_{X}A$ contains the origin, ${\mathrm{Int}}_{X}A$ will also be balanced. If $W$ is convex then $A$ is convex and balanced and thus the same is true of ${\mathrm{Int}}_{X}A.$ $◼$

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 $(X,p_D)$ into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples $rD$ as $r$ ranges over $\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \}$ (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 $D:=\bigcap _{|u|=1}uW;$ if in addition $D$ does not contain any non-trivial vector subspace then $p_D$ will be a norm and $(X,p_D)$ 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

See also: Topological vector space § 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 $|c|\le |d|$ then $cB\subseteq dB$ and $cB=|c|B.$ Thus if $c$ and $d$ are any scalars then $(cB)\cap (dB)=\underset{}{\min }\{|c|,|d|\}B.$
• $B$ 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 $\mathrm{span}B$ then $B\cap Y$ is either $\{0\}$ or else it is absorbing in $Y.$
• for any $x\in X,$ if $B\cap \mathrm{span}x$ contains more than one point then it is a convex and balanced neighborhood of $0$ in the 1-dimensional vector space $\mathrm{span}x$ 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 ${𝒮}$ of subsets of $X,$ $$\mathrm{bal}\left(\bigcup _{S\in {𝒮}}S\right)=\bigcup _{S\in {𝒮}}\mathrm{bal}S\text{ and }\mathrm{balcore}\left(\bigcap _{S\in {𝒮}}S\right)=\bigcap _{S\in {𝒮}}\mathrm{balcore}S.$$ 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 $c,$ $\mathrm{bal}(cS)=c\mathrm{bal}S=|c|\mathrm{bal}S.$ For any scalar $c\ne 0,$ $\mathrm{balcore}(cS)=c\mathrm{balcore}S=|c|\mathrm{balcore}S.$ This equality holds for $c=0$ if and only if $S\subseteq \{0\}.$ Thus if $0\in S$ or $S=\varnothing$ then $$\mathrm{balcore}(cS)=c\mathrm{balcore}S=|c|\mathrm{balcore}S$$ for every scalar $c.$

Related notions

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

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

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.

See also

• Absolutely convex set – Convex and balanced set
• Absorbing set – Set that can be "inflated" to reach any point
• Bounded set (topological vector space) – Generalization of boundedness
• Convex set – In geometry, set whose intersection with every line is a single line segment
• Star domain – Property of point sets in Euclidean spaces
• Symmetric set – Property of group subsets (mathematics)
• Topological vector space – Vector space with a notion of nearness

References

Proofs

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