Prove that any $T_1$ topological vector space is regular

general-topologysolution-verificationtopological-vector-spaces

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?

Best Answer

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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