Geometry of Topological Vector Spaces

Seperation

Balanced and Absorbing Sets

Balanced Set

A set in a real or complex vector space is called balanced if for every and every scalar with , we have .

Proposition

Suppose is a balanced set. Then for all scalars , with .

Proof As , for any , we have So .

Absorbing Set

A set in a real or complex vector space is called absorbing if for every , there exists a such that for all .

It is quite easy to see that absorbing sets, balanced sets, and convex sets (containing ) are similar and related. However, it turns out that they are completely different, satisfying one or two of the properties does not imply the other. Here are some classical examples that can help to gain some intuition.

Remark

Every balanced set in (as a complex vector space) is convex. However, this is not true in higher dimensional, nor when regarding as a two-dimensional real vector space.

	Real/Complex	Absorbing	Balanced	Convex
Open unit disk in		✅	✅	✅
Star shape:	Real	✅	✅	🚫
Closed unit disk with a point:		✅	🚫	🚫
	Complex	🚫	✅	🚫
Unit cross:	Real	🚫	✅	🚫
Unit interval in	Complex	🚫	🚫	✅

Proposition

Let be a topological vector space, and be an open neighborhood of , then for any sequence of real numbers with . In particular, every open neighbourhood of is absorbing.

Proof Let , consider . Note that . Define , , then is continuous and is open. Since diverges, there exists such that for all . Thus for all .

Proposition

Let be a topological vector space, and be an open neighborhood of , then for any sequence of real numbers with . In particular, every open neighbourhood of is absorbing. Every topological vector space has a basis of balanced neighborhoods.

Proof It suffices to show a balanced basis at the origin.

Boundedness

Boundedness in Topological Vector Spaces

A subset of a topological vector space is bounded if for all open neighbourhood of the origin, there exists a scalar such that for all .

Lemma

Translation and scalar multiplication of bounded sets are bounded. More generally, if is bounded and is bounded, then is bounded for all scalars .

Proof It suffices to only prove the general case, and assume that . Let be an open neighborhood of . By continuity of addition, we can find an open neighbourhood of such that . Since is bounded, there exists such that for all . Similarly, there exists such that for all , that is, for all . Let , then for all , we have So is bounded.

Form the lemma, we immediately have

Proposition

in a topological space is bounded if and only if it is bounded by neighbourhoods at the origin, thus all points in .

Proposition

Finite union of bounded sets is bounded.

Proof Suppose is some finite index set and is a bounded set. Then for any open neighbourhood of , there exists a scalar such that for all . Let , then for all , we have . So is bounded.