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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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What do you mean by "the local base"? I have only ever heard "local base" applied to a system of neighborhoods around a single point. Please also include any separation axioms you intended to assume. –  rschwieb Jun 7 '12 at 13:15
    
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? –  Vobo Jun 7 '12 at 13:24

2 Answers 2

up vote 4 down vote accepted

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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I would want you to check if my argument is correct for the fact that closure of $K+V$ doesn't intersect $C$. My argument is that if closure of $K+V$ intersects $C$ then there exists some open set of $K$ ie $x+V_x$ which intersects with $C$ contradicting the fact that $K\cap C =\varnothing$. Am i right ? –  Theorem Jun 8 '12 at 17:20
    
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. –  no identity Jun 8 '12 at 17:26

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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Isn't the regularity of a TVS usually proved by checking that a TVS has a base of closed neighbourhoods? –  Vobo Jun 7 '12 at 14:26
    
@Vobo It could be: it has been years since I learned about these. –  rschwieb Jun 7 '12 at 14:28

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