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Proposition

For a $T_1$ space $(X,\mathscr T)$ the following three conditions are equivalent:

  1. X is regular;
  2. for any $x\in X$ and any $U$ containing it there exist an open neighborhood of $x$ such that $V\subseteq\overline V\subseteq U$;
  3. any $x\in X$ has a local base whose element are closed.

Theorem

Suppose $K$ and $C$ are disjoint subsets of a topological vector space X such that K is compact and C is closed. so there exist a symmetric (open) neighborhood $V$ of $0$ such that $$ (K+V)\cap(C+K)=\emptyset $$

Corollary

Let be $X$ a topological vector space. If $\mathscr B(0)$ is a local (open) base centered at $0$ then any its element contains the closure of another one element.

So knowing these results I ask to me if any $T_1$ topological vector space is regular too and so I arranged the following arguments. If $U$ is an open set containing $x$ then $U-x$ is an open set (any raslation is an homeomorphism!) containing $0$ and so there exist an open neighborhood $V$ of $0$ whose closure is contained $U-x$ so that $V+x$ is an open set containing $x$ whose closure is contained in $U$ and thus the statement follows immediately by the proposition mentioned above. So is the argument true and are my arguments correct? Could someone help me, please?

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  • $\begingroup$ That any $T_1$ topological vector space is regular is a very basic theorem which can be found in many books. In particular Rudin's FA has a proof. $\endgroup$ Jan 5 at 12:11
  • $\begingroup$ I am reading Rudin's book but strangerly using the theorem I mentioned he proves only that any $T_1$ topological vector space is hausdorff so that I thought to ask these question because I suppose that it could be false if Rudin did not prove this, that's all. $\endgroup$ Jan 5 at 12:12
  • $\begingroup$ Anyway are my arguments correct? $\endgroup$ Jan 5 at 12:13
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    $\begingroup$ Higher level argument (but Rudin doesn't cover it in this way): a TVS is a uniform space (being a topological group) and a uniformisable $T_1$ topology is Tikhonov (or completely regualr or $T_{3\frac12}$, hence $T_3$ a fortiori. $\endgroup$ Jan 5 at 21:19
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Your argument is correct. You could also apply the theorem directly: if $x\in U\subseteq X$, where $U$ is open, then $\{x\}$ is compact, and $C=X\setminus U$ is closed and disjoint from $\{x\}$, so there is an open nbhd $V$ of $0$ such that $(x+V)\cap(C+V)=\varnothing$. But then $C+V$ is an open nbhd of $C$, so $F=X\setminus(C+V)$ is a closed subset of $U$, and $x+V$ is an open nbhd of $x$ such that $\operatorname{cl}(x+V)\subseteq F\subseteq U$.

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  • $\begingroup$ Thanks for the answer. Could I ask your assistance here too, please? $\endgroup$ Jan 5 at 20:01

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