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 $$a S \subseteq S$$ 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 $$a S = \{a s : s \in S\}$$ and $$B S = \{b s : b \in B, s \in S\}$$ 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  or balanced if it satisfies any of the following equivalent conditions:  Definition: $$a s \in S$$ for all $$s \in S$$ and all scalars $$a$$ satisfying $$|a| \leq 1.$$ $$a S \subseteq S$$ 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 $$s \in S,$$ $$S \cap \mathbb{K} s = B_{\leq 1} (S \cap \mathbb{K} s).$$ 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). 
 * $$\mathbb{K} s = \operatorname{span} \{s\}$$ 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).</li>
 * 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. </li>

If $$S$$ is a convex set then this list may be extended to include: <li>$$a S \subseteq S$$ for all scalars $$a$$ satisfying $$|a| = 1.$$</li> </ol>

If $$\mathbb{K} = \R$$ then this list may be extended to include: <li>$$S$$ is symmetric (meaning $$- S = S$$) and $$[0, 1) S \subseteq S.$$</li> </ol>

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

The  of a subset $$S$$ of $$X,$$ denoted by $$\operatorname{bal} S,$$ is defined in any of the following equivalent ways: <ol> <li>Definition: $$\operatorname{bal} S$$ is the smallest (with respect to $$\,\subseteq\,$$) balanced subset of $$X$$ containing $$S.$$</li> <li>$$\operatorname{bal} S$$ is the intersection of all balanced sets containing $$S.$$</li> <li>$$\operatorname{bal} S = \bigcup_{|a| \leq 1} (a S).$$</li> <li>$$\operatorname{bal} S = B_{\leq 1} S.$$</li> </ol>

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

The  of a subset $$S$$ of $$X,$$ denoted by $$\operatorname{balcore} S,$$ is defined in any of the following equivalent ways: <ol> <li>Definition: $$\operatorname{balcore} S$$ is the largest (with respect to $$\,\subseteq\,$$) balanced subset of $$S.$$</li> <li>$$\operatorname{balcore} S$$ is the union of all balanced subsets of $$S.$$</li> <li>$$\operatorname{balcore} S = \varnothing$$ if $$0 \not\in S$$ while $$\operatorname{balcore} S = \bigcap_{|a| \geq 1} (a S)$$ if $$0 \in S.$$</li> </ol>

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 $$\operatorname{Int}_X U ~\subseteq~ B_1 U ~=~ \bigcup_{0 < |a| < 1} a U ~\subseteq~ U.$$

Balanced sets in $$\R$$ and $$\Complex$$

Let $$\mathbb{K}$$ be the field real numbers $$\R$$ or complex numbers $$\Complex,$$ 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 $$\mathbb{K} := \Complex$$ is the field of complex numbers then $$X = \mathbb{K} = \Complex$$ is a 1-dimensional complex vector space whereas if $$\mathbb{K} := \R$$ then $$X = \mathbb{K} = \R$$ is a 1-dimensional real vector space.

The balanced subsets of $$X = \mathbb{K}$$ are exactly the following: <ol> <li>$$\varnothing$$</li> <li>$$X$$</li> <li>$$\{0\}$$</li> <li>$$\{x \in X : |x| < r\}$$ for some real $$r > 0$$</li> <li>$$\{x \in X : |x| \leq r\}$$ for some real $$r > 0.$$</li> </ol> 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 $$\Complex$$ 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, $$\Complex$$ and $$\R^2$$ are entirely different as far as scalar multiplication is concerned.

Balanced sets in $$\R^2$$

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

If $$x_0 \in X = \R^2$$ is non-zero, and $$L := \R x_0,$$ 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 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 $$\R^n$$ for any integer $$n \geq 1.$$

Let $$B \subseteq \R^2$$ 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 $$\operatorname{span} B = \R^2$$ 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 $$X := \R^2$$ 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 $$x y = 1$$ in $$X = \R^2.$$

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 $$S := [-1, 1] \times \{1\},$$ which is a horizontal closed line segment lying above the $$x-$$axis in $$X := \R^2.$$ 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 $$\mathbb{K}$$).

<ul> <li>The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.</li> <li>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).</li> <li>Arbitrary unions of balanced sets are balanced, and the same is true of arbitrary intersections of balanced sets.</li> <li>Scalar multiples and (finite) Minkowski sums of balanced sets are again balanced.</li> <li>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.</li> </ul>

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, $$\left\{(z, w) \in \Complex^2 : |z| \leq |w|\right\}$$ is a balanced subset of $$X = \Complex^2$$ 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 $$X := \Reals^2$$ 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 $$\bigcap_{|u|=1} u W \subseteq W$$ 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 $$\bigcap_{|u|=1} u W$$ will be a convex neighborhood of the origin and so its topological interior will be a balanced convex  neighborhood of the origin.

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Suppose that $$W$$ is a convex and absorbing subset of $$X.$$ Then $$D := \bigcap_{|u|=1} u W$$ will be convex balanced absorbing subset of $$X,$$ which guarantees that the Minkowski functional $$p_D : X \to \R$$ 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 $$r D$$ 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 $$D := {\textstyle\bigcap\limits_{|u|=1}} u W;$$ 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.

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: <ul> <li>for any scalars $$c$$ and $$d,$$ if $$|c| \leq |d|$$ then $$c B \subseteq d B$$ and $$c B = |c| B.$$ Thus if $$c$$ and $$d$$ are any scalars then $$(c B) \cap (d B) = \min_{} \{|c|, |d|\} B.$$</li> <li>$$B$$ is absorbing in $$X$$ if and only if for all $$x \in X,$$ there exists $$r > 0$$ such that $$x \in r B.$$</li> <li>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 $$Y.$$</li> <li>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 $$B \cap \R x$$ is a convex balanced subset of the real vector space $$\R x$$ that contains the origin.</li> </ul>

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 if it satisfies any of the following equivalent conditions: <ol> <li>$$p(a x) \leq p(x)$$ whenever $$a$$ is a scalar satisfying $$|a| \leq 1$$ and $$x \in X.$$</li> <li>$$p(a x) \leq p(b x)$$ whenever $$a$$ and $$b$$ are scalars satisfying $$|a| \leq |b|$$ and $$x \in X.$$</li> <li>$$\{x \in X : p(x) \leq t\}$$ is a balanced set for every non-negative real $$t \geq 0.$$</li> </ol> If $$p$$ is a balanced function then $$p(a x) = p(|a| x)$$ for every scalar $$a$$ and vector $$x \in X;$$ so in particular, $$p(u x) = p(x)$$ 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 $$p : X \to \R$$ is a seminorm if and only if it is a balanced sublinear function.