In linear algebra and related areas of mathematics a **balanced set**, **circled set** or **disk** in a vector space (over a field with an absolute value function ) is a set such that for all scalars satisfying

The **balanced hull** or **balanced envelope** of a set is the smallest balanced set containing
The **balanced core** of a set is the largest balanced set contained in

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.

Let be a vector space over the field of real or complex numbers.

**Notation**

If is a set, is a scalar, and then let and and for any let

denote, respectively, the

**Balanced set**

A subset of is called a *balanced set* or *balanced* if it satisfies any of the following equivalent conditions:

*Definition*: for all and all scalars satisfying- for all scalars satisfying
- (where ).
^{[1]}- For every
- is a (if ) or (if ) dimensional vector subspace of
- If then the above equality becomes which is exactly the previous condition for a set to be balanced. Thus, is balanced if and only if for every is a balanced set (according to any of the previous defining conditions).

- For every 1-dimensional vector subspace of is a balanced set (according to any defining condition other than this one).
- For every there exists some such that or
- is a balanced subset of (according to any defining condition of "balanced" other than this one).
- Thus is a balanced subset of if and only if it is balanced subset of every (equivalently, of some) vector space over the field that contains So assuming that the field is clear from context, this justifies writing " is balanced" without mentioning any vector space.
^{[note 1]}

- Thus is a balanced subset of if and only if it is balanced subset of every (equivalently, of some) vector space over the field that contains So assuming that the field is clear from context, this justifies writing " is balanced" without mentioning any vector space.

If is a convex set then this list may be extended to include:

- for all scalars satisfying
^{[2]}

If then this list may be extended to include:

- is symmetric (meaning ) and

The *balanced hull* of a subset of denoted by is defined in any of the following equivalent ways:

*Definition*: is the smallest (with respect to ) balanced subset of containing- is the intersection of all balanced sets containing
^{[1]}

The *balanced core* of a subset of denoted by is defined in any of the following equivalent ways:

*Definition*: is the largest (with respect to ) balanced subset of- is the union of all balanced subsets of
- if while if

The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, 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 is a seminorm (or norm) on a vector space then for any constant the set is balanced.

If is any subset and then is a balanced set. In particular, if is any balanced neighborhood of the origin in a topological vector space then

**Balanced sets in and **

Let be the field real numbers or complex numbers let denote the absolute value on and let denotes the vector space over 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:^{[3]}

- for some real
- for some real

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 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, and are entirely different as far as scalar multiplication is concerned.

**Balanced sets in **

Throughout, let (so is a vector space over ) and let is the closed unit ball in centered at the origin.

If is non-zero, and then the set is a closed, symmetric, and balanced neighborhood of the origin in More generally, if is *any* closed subset of such that then is a closed, symmetric, and balanced neighborhood of the origin in This example can be generalized to for any integer

Let be the union of the line segment between the points and and the line segment between and Then is balanced but not convex. Nor is is absorbing (despite the fact that is the entire vector space).

For every let be any positive real number and let be the (open or closed) line segment in between the points and Then the set 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 in

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 which is a horizontal closed line segment lying above the axis in The balanced hull is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles and where and is the filled triangle whose vertices are the origin together with the endpoints of (said differently, is the convex hull of while is the convex hull of ).

A set 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 ).

- 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 is a linear map and and are balanced sets, then and are balanced sets.

In any topological vector space, the closure of a balanced set is balanced.^{[5]} The union of the origin 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, 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.^{[6]} Similarly for real vector spaces, if denotes the convex hull of and (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 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 ) whenever this is true of It will be a balanced set if is a star shaped at the origin,^{[note 2]} which is true, for instance, when is convex and contains In particular, if is a convex neighborhood of the origin then 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]}

Let and define (where denotes elements of the field of scalars). Taking shows that If is convex then so is (since an intersection of convex sets is convex) and thus so is 's interior. If then

and thus If is star shaped at the origin

thus proving that is balanced.
If is convex and contains the origin then it is star shaped at the origin and so will be balanced.

Now suppose is a neighborhood of the origin in Since scalar multiplication (defined by ) is continuous at the origin and there exists some basic open neighborhood (where and ) of the origin in the product topology on such that the set is balanced and it is also open because it may be written as

where is an open neighborhood of the origin whenever
Finally,

shows that is also a neighborhood of the origin.
If is balanced then because its interior contains the origin, will also be balanced.
If is convex then is convex and balanced and thus the same is true of

Suppose that is a convex and absorbing subset of Then will be convex balanced absorbing subset of which guarantees that the Minkowski functional of will be a seminorm on thereby making into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples as ranges over (or over any other set of non-zero scalars having as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If is a topological vector space and if this convex absorbing subset is also a bounded subset of then the same will be true of the absorbing disk if in addition does not contain any non-trivial vector subspace then will be a norm and will form what is known as an auxiliary normed space.^{[7]} If this normed space is a Banach space then is called a *Banach disk*.

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 is a balanced subset of then:

- for any scalars and if then and Thus if and are any scalars then
- is absorbing in if and only if for all there exists such that
^{[2]} - for any 1-dimensional vector subspace of the set is convex and balanced. If is not empty and if is a 1-dimensional vector subspace of then is either or else it is absorbing in
- for any if contains more than one point then it is a convex and balanced neighborhood of in the 1-dimensional vector space when this space is endowed with the Hausdorff Euclidean topology; and the set is a convex balanced subset of the real vector space that contains the origin.

**Properties of balanced hulls and balanced cores**

For any collection of subsets of

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open.
If is a Hausdorff topological vector space and if is a compact subset of then the balanced hull of 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 and any scalar

For any scalar This equality holds for if and only if Thus if or then

for every scalar

A function on a real or complex vector space is said to be a *balanced function* if it satisfies any of the following equivalent conditions:^{[9]}

- whenever is a scalar satisfying and
- whenever and are scalars satisfying and
- is a balanced set for every non-negative real

If is a balanced function then for every scalar and vector
so in particular, for every unit length scalar (satisfying ) and every ^{[9]}
Using shows that every balanced function is a symmetric function.

A real-valued function is a seminorm if and only if it is a balanced sublinear function.