I would like to prove that if $B$ is local base for a topological vector space $X$, then every member of $B$ contains the closure of some member of $B$.
I would appreciate if somebody can guide me through this problem. I am still facing problems in understanding various primary concepts.
On
Have you learned that (a $T_0$) topological vector space is regular?
If so, then you can use the following lemma to prove your statement.
Lemma: A topological space $X$ is regular iff for every $x\in X$ and open set $\mathcal{O}$ containing $x$, there exists an open set $\mathcal{U}$ such that $x\in\mathcal{U}\subseteq \overline{\mathcal{U}}\subseteq \mathcal{O}$.
Since you read Ridin's functional analysis I will refer to its theorems.
Let $U\in B$ be some neighborhood of zero and element of the base $B$. Consider compact $K=\{0\}$, and closed set $C=X\setminus U$. By corollary of theorem 1.10 there exist neighborhood of zero $V$ such that $\overline{(K+V)}\cap (C+V)=\varnothing$.
Assume that, $\overline{(K+V)}\cap C\neq\varnothing$ , then there exist $c\in C$ such that $c\in \overline{(K+V)}$. Since $C\subset C+V$, then we have $c\in C+V$ such that $c\in \overline{(K+V)}$. This means that $\overline{(K+V)}\cap(C+V)\neq\varnothing$. Contradiction, hence $\overline{(K+V)}\cap C=\varnothing$.
Since $\overline{(K+V)}\cap C=\varnothing$ then $\overline{K+V}\subset X\setminus C=U$. Since $K=\{0\}$, then $\overline{(K+V)}=\overline{V}$, so we get $\overline{V}\subset U$. Since $B$ is a local base of zero, then there exist $W\in B$ such that $W\subset V$. As the consequence $\overline{W}\subset\overline{V}\subset U$. Thus for each $U\in B$ we found $W\in B$ such that $\overline{W}\subset U$.