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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.

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    For a 0-neighbourhood $U$, $x \in \overline{U}$ implies $ (x+U)\cap U \neq \emptyset$, i.e. $x+u_1 = u_2$. Can you proceed from here?2012-06-07

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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 $\overline{W}\subset U$.

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    If $x+V_x$ intersects $C$ then the point of intersection $p$ not necessary in $K$. So you proof is not correct. I'edit my answer, to get things more clear.2012-06-08
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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 $x\in\mathcal{U}\subseteq \overline{\mathcal{U}}\subseteq \mathcal{O}$.

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    @Vobo It could be: it has bee$n$ years since I learned about these.2012-06-07