Review of Topological Vector Spaces

# Review of Topological Vector Spaces

 Results on Topological Vector Spaces (1) For each $a \in E$, $x \mapsto a + x$ is a homeomorphism of $E$ onto $E$. (2) For each $\alpha \in \mathbf{F}$, $\lambda \neq 0$, $x \mapsto \alpha x$ is a homeomorphism of $E$ onto $E$.
• On the Bases of Neighbourhoods for a Point in a Topological Vector Space we looked at bases of neighbourhoods of points in a topological vector space. Recall that if $E$ is a topological vector space and if $\mathcal U$ is a base of neighbourhoods of the origin then $\mathcal U + a$ is a base of neighbourhoods of the point $a$. If $\mathcal U$ is a base of neighbourhoods of the origin then the following properties are satisfied:
 Properties of a Base of Neighbourhoods of the Origin for a Topological Vector Space (1) Every $U \in \mathcal U$ is an absorbent set. (2) For each $U \in \mathcal U$ there exists a balanced $V \in \mathcal U$ with $V \subseteq V + V \subseteq U$. (3) For every convex $U \in \mathcal U$ there is an absolutely convex $W \in \mathcal U$ with $W \subseteq U$.
 Closures in a Topological Vector Space (1) If $E$ is a topological vector space and $A \subseteq E$ is convex then $\overline{A}$ is convex. (2) If $E$ is a topological vector space and $A \subseteq E$ is balanced then $\overline{A}$ is balanced. (3) If $E$ is a topological vector space and $A \subseteq E$ is absolutely convex then $\overline{A}$ is absolutely convex.

If $E$ is a topological vector space then $E$ has a base of closed and balanced neighbourhoods of the origin.

A topological vector space $E$ is Hausdorff if and only if $\displaystyle{\bigcap_{U \in \mathcal U} U = \{ o \}}$ where $\mathcal U$ is a base of neighbourhoods of the origin.

If $E$ is a locally convex topological vector space then $E$ has a base of closed absolutely convex and absorbent neighbourhoods $\mathcal U$ of the origin with the property that if $U, V \in \mathcal U$ then there exists a $W \in \mathcal U$ with $W \subseteq U \cap V$ and such that if $U \in \mathcal U$ and $\alpha \in \mathbf{F}$ with $\alpha \neq 0$ then $\alpha U \in \mathcal U$.