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 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 then let and and for any let denote, respectively, the open ball and the closed ball of radius r in the scalar field \mathbb{K} centered at 0 where and Every balanced subset of the field \mathbb{K} is of the form B_{\leq r} or B_r for some Balanced set A subset S of X is called a or balanced if it satisfies any of the following equivalent conditions:

<ol> <li>Definition: a s \in S for all s \in S and all scalars a satisfying |a| \leq 1.</li> <li> for all scalars a satisfying |a| \leq 1.</li> <li> (where ).</li> <li></li> <li>For every s \in S, <li>For every 1-dimensional vector subspace Y of S \cap Y is a balanced set (according to any defining condition other than this one).</li> <li>For every s \in S, there exists some such that or </li> <li>S is a balanced subset of (according to any defining condition of "balanced" other than this one). </ol> If S is a [convex set](https://bliptext.com/articles/convex-set) then this list may be extended to include: <li> for all scalars a satisfying |a| = 1.</li> </ol> If then this list may be extended to include: <li>S is [symmetric](https://bliptext.com/articles/symmetric-set) (meaning - S = S) and </li> </ol>

Balanced hull

The of a subset S of X, denoted by is defined in any of the following equivalent ways:

<ol> <li>Definition: is the smallest (with respect to ) balanced subset of X containing S.</li> <li> is the [intersection](https://bliptext.com/articles/intersection-set-theory) of all balanced sets containing S.</li> <li></li> <li></li> </ol>

Balanced core

The of a subset S of X, denoted by is defined in any of the following equivalent ways:

<ol> <li>Definition: is the largest (with respect to ) balanced subset of S.</li> <li> is the union of all balanced subsets of S.</li> <li> if 0 \not\in S while 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 is balanced. If is any subset and then B_1 S is a balanced set. In particular, if is any balanced neighborhood of the origin in a topological vector space X then 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 denotes the vector space over \mathbb{K}. So for example, if is the field of complex numbers then is a 1-dimensional complex vector space whereas if then is a 1-dimensional real vector space. The balanced subsets of are exactly the following:

<ol> <li>\varnothing</li> <li>X</li> <li>\{0\}</li> <li> for some real r > 0</li> <li> for some real r > 0.</li> </ol> Consequently, both the [balanced core](https://bliptext.com/articles/) and the [balanced hull](https://bliptext.com/articles/) 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](https://bliptext.com/articles/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 is non-zero, and then the set is a closed, symmetric, and balanced neighborhood of the origin in X. More generally, if C is closed subset of X such that then 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 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 is the entire vector space). For every 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 and Then the set is a balanced and absorbing set but it is not necessarily convex. The [balanced hull](https://bliptext.com/articles/) 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](https://bliptext.com/articles/) 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 which is a horizontal closed line segment lying above the x-axis in X := \R^2. The balanced hull is a non-convex subset that is "[hour glass](https://bliptext.com/articles/hour-glass) shaped" and equal to the union of two closed and filled [isosceles triangles](https://bliptext.com/articles/isosceles-triangle) 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](https://bliptext.com/articles/convex-hull) of while T_2 is the [convex hull](https://bliptext.com/articles/convex-hull) of ).

Sufficient conditions

A set T is balanced if and only if it is equal to its balanced hull or to its balanced core in which case all three of these sets are equal: 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](https://bliptext.com/articles/compact-space) (respectively, [totally bounded](https://bliptext.com/articles/totally-[bounded](https://bliptext.com/articles/bounded-set-topological-vector-space)), [bounded](https://bliptext.com/articles/bounded-set-topological-vector-space)) set has the same property.</li> <li>The convex hull of a balanced set is convex and balanced (that is, it is [absolutely convex](https://bliptext.com/articles/absolutely-convex-set)). However, the balanced hull of a convex set may fail to be convex (a counter-example is given above).</li> <li>Arbitrary [unions](https://bliptext.com/articles/union-set-theory) of balanced sets are balanced, and the same is true of arbitrary [intersections](https://bliptext.com/articles/intersection-set-theory) of balanced sets.</li> <li>Scalar multiples and (finite) [Minkowski sums](https://bliptext.com/articles/minkowski-sum) of balanced sets are again balanced.</li> <li>Images and preimages of balanced sets under [linear maps](https://bliptext.com/articles/linear-map) are again balanced. Explicitly, if L : X \to Y is a linear map and and 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, is a balanced subset of that contains the origin 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 is an (hour glass shaped) balanced subset of whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set 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 the symmetric set 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 will be a convex neighborhood of the origin and so its topological interior will be a balanced convex neighborhood of the origin. Suppose that W is a convex and absorbing subset of X. Then will be convex balanced absorbing subset of X, which guarantees that the Minkowski functional of D will be a seminorm on X, thereby making into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples r D as r ranges over (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 if in addition D does not contain any non-trivial vector subspace then p_D will be a norm and 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 then and Thus if c and d are any scalars then </li> <li>B is [absorbing](https://bliptext.com/articles/absorbing-set) 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 then B \cap Y is either \{0\} or else it is [absorbing](https://bliptext.com/articles/absorbing-set) in Y.</li> <li>for any x \in X, if contains more than one point then it is a convex and balanced neighborhood of 0 in the 1-dimensional vector space when this space is endowed with the [Hausdorff](https://bliptext.com/articles/hausdorff-space) [Euclidean topology](https://bliptext.com/articles/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, In any topological vector space, the [balanced hull](https://bliptext.com/articles/) of any open neighborhood of the origin is again open. If X is a [Hausdorff](https://bliptext.com/articles/hausdorff-space) [topological vector space](https://bliptext.com/articles/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](https://bliptext.com/articles/absorbing-set), a neighborhood of the origin) then the same is true of its balanced core. For any subset and any scalar c, For any scalar c \neq 0, This equality holds for c = 0 if and only if Thus if 0 \in S or then for every scalar c.

Related notions

A function on a real or complex vector space is said to be a if it satisfies any of the following equivalent conditions:

<ol> <li> whenever a is a scalar satisfying |a| \leq 1 and x \in X.</li> <li> whenever a and b are scalars satisfying and x \in X.</li> <li> is a balanced set for every non-negative real t \geq 0.</li> </ol> If p is a balanced function then for every scalar a and vector x \in X; so in particular, for every [unit length](https://bliptext.com/articles/unit-length) scalar u (satisfying |u| = 1) and every x \in X. Using u := -1 shows that every balanced function is a [symmetric function](https://bliptext.com/articles/symmetric-function). A real-valued function is a [seminorm](https://bliptext.com/articles/seminorm) if and only if it is a balanced [sublinear function](https://bliptext.com/articles/sublinear-function).

Sources