T2 topological vector space is T3

Question

I trying to understand that if a TVS V is Hausdorff, that we then can separate closed sets and points by opens.

In the accepted answer of Every Topological Vector Space is Regular, I have two questions:

(1) Why can one select opens $O,O'$ such that $Y\times 0\subseteq O \times O' \subseteq f^{-1}(U)$? Somehow this uses continuity. I can see that for every $y\in Y$ we can select opens with $(y,0)\in O\times O' \subseteq f^{-1}(U)$ but why can we choose these to work for all $y\in Y$ at once?

(2) Where does this proof use that $Y$ is closed?

If someone can give an alternative proof using elementary tools (just the definitions and basic topology), that will equally answer my question. Just trying to understand why this result holds.

Thanks.

Answer 1

I am having some issues with the linked question and answers, so I'll explain my own approach instead.

Here's a trick I like: topological groups, vector spaces, etc. are a little weird to work at. But these types of problems actually depend only on the uniform structure of the given topological spaces.


I like metric spaces, though.


So what would the similar result be for metric spaces?

Let $(M,d)$ be a metric space. Then for every closed $Y\subseteq M$ and $x\not\in Y$, there exist disjoint open sets $U,U'$ such that $x\in U$ and $Y\subseteq U'$.

This is easy to do geometrically: Since $x\not\in Y$ and $Y$ is closed, then the distance from $x$ to $Y$ is positive. Let $d=d(x,Y)=\inf\left\{d(x,y):y\in Y\right\}>0$ be that distance. Then $$B_{d/2}(x)=\left\{z\in M:d(x,z)<d/2\right\}$$ and $$B_{d/2}(Y)=\left\{z\in M:d(z,Y)<d/2\right\}$$ have precisely the property that we need.

Ok, so what was the main point of the argument? We basically did the following steps:

1. Find an open set containing $x$ which does not intersect $Y$: $B_d(x)$ does that;

2. Take another open subset which has half of the diameter of the original open set: $B_{d/2}(x)$ does that

3. Translate that small open set to $Y$: $B_{d/2}(Y)$ works;

4. Done.

The same procedure can be done in topological vector spaces, topological groups, etc. For step 2., you should use the following facts:

• If $x$ is a point in a topological vector space, then the open sets around $x$ are precisely those of the form $x+O$, where $O$ is an open set around $0$;

• If $O$ is an open set around $0$, then there exists another open set $O'$ around $0$ such that $O'+O'\subseteq O$ (because addition is continuous).

• Then $x+O'$ is an open set around $x$ which has "intuitive diameter half of that of $x+O$;