# Prove that any $T_1$ topological vector space is regular

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?

• 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. Jan 5 at 12:11
• 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. Jan 5 at 12:12
• Anyway are my arguments correct? Jan 5 at 12:13
• 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. Jan 5 at 21:19

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 $$\operatorname{cl}(x+V)\subseteq F\subseteq U$$.