# Every Topological Vector Space is Regular

Let $$V$$ be an abstract ($$T_0$$) topological vector space over topological field $$K$$ (We may assume that $$K = \mathbb{C}$$ or $$K = \mathbb{R}$$ for simplicity). This means that the only thing that we are allowed to use is that scalar multiplication and addition are jointly continuous in both arguments (and also the $$T_0$$ properties). The problem is to prove:

For each point $$x$$ and closed set $$Y$$ such that $$x \not \in Y$$ there are exist disjoint open neighborhoods $$O$$ and $$O'$$ such that $$x \in O$$ and $$Y \subset O'$$.

I will write in the sequel for such sets $$O \in \mathcal{U}_V(x),O' \in \mathcal{U}_V(Y)$$.

If $$U \in \mathcal{U}_V(y)$$ then we can squeeze it by factor of $$r \in K$$ $$r \ast_y U = \big(rU + \{(1 - r)y\}\big) \cap U$$ which is open neighborhood of $$y$$.

for all $$y \in Y$$ there must exist $$U_y \in \mathcal{U}_V(y)$$ such that $$x \not \in U_y$$, otherwise $$x \in \lim_{n \to \infty} y$$ and so $$x \in Y$$ which is not true. Then, $$U = \bigcup_{y \in Y} U_y \in \mathcal{U}_V(Y)$$ do not contain $$x$$. In case $$x \not \in \delta U$$ just let $$O' = U$$. Otherwise, squeeze all open sets by some values $$r_y$$: $$O' = \bigcup_{y \in Y} r_y \ast_y U_y$$ Then $$O = \Big(\overline{O'}\Big)^\complement$$. How to select $$r_y$$? Will this approach work?

• If $Y$ is not compact, that approach may not work. It's anyway easier to show that every neighbourhood of a point contains a closed neighbourhood (i.e. topological vector spaces are $T_3$; actually, since they are uniform spaces, they are $T_{3\frac{1}{2}}$). For $U_x$, there is a neighbourhood $V$ of $0$ with $x + V + V \subset U_x$. Jun 22, 2017 at 19:29
• Usually additional axioms are included, such as {0} is closed. Otherwise the space could have the indiscrete topology. Jun 22, 2017 at 19:35
• Side remark: Only the fact that we're dealing with a topological group is needed. Scalar multiplication can be completely ignored. Jun 22, 2017 at 19:38
• What you're trying to show is implied by the fact that every topological group is regular. Jun 22, 2017 at 20:08

This is a version of the proof redrawn from topological groups:

$$f(x,y) = x - y$$ is continues function.

Without loss of generality we may assume that $$x = 0$$.

Then, select $$U \in \mathcal{U}_V(Y)$$ such that $$0 \not \in U$$, then where must exist two open sets $$O$$ and $$O'$$ such that $$Y \times \{ 0\} \subset O' \times O \subset f^{-1}$$(U) as $$y - 0 = y$$ for all $$y \in Y$$.

Now we show that $$A =O' \cap (V \setminus U) + O = \emptyset$$.

For every $$y \in A$$, there is a representation $$y = v +z$$ for some $$z \not \in U$$ and $$v\in O$$ . We also know that $$y \in O'$$. This means that $$y - v \in U$$ but we also know that $$y - v = z \not \in U$$, a contradiction. This shows that $$A = \emptyset$$

If $$U$$ was constructed as in the question above, then $$O \subset (V\setminus U) + O$$ as $$0 \not \in U$$ . Thus, $$O \cap O' = \emptyset$$

$$\square$$

This answer was written in order to close the question as it was essentially answered in the comments.

• Don't we need the topology of the space $V$ to be, necessarily, Hausdorff? (Or, $V$ is a $T_1$ space, or, even just $\{0\}$ is closed in $V$?) If so, in the proof, in which step we use the fact that the topology of $V$ is Hausdorff? Mar 12, 2021 at 14:46
• @serenus Yes the space must be T2 as we proved that the space is T3 in fact. The Correct statement of the theorem is that every T0 Topological group (or vector space) is T3.5, which is also true for any uniform space. Non T0 VTS do exists, but there is a fact that every suxh space $V$ can be represented as $V = C \oplus R$, where $C$ is codescrete and $R$ is regular. In such a space we cant' assert that $0 \not \in U$. Otherwise, it is easy to show that $T0 \Rightarrow T1 \Rightarrow$ such $U$ exists. Mar 12, 2021 at 15:59
• Mar 12, 2021 at 16:25
• In your proof you state that: "Then, select $U\in\mathcal{U}_V(Y)$ such that $0\notin U$ ...". How can we get such a neighborhood $U$ not containing $0$? Is it because $V$ is a $T_0$ space? Mar 12, 2021 at 20:42
• @serenus Right. The definition of $T_0$ says that for every pair of points $x,y$ there is is the neighbourhood of one of this points $U$ which does not contain another. But by continuous addition we can easly transfer neighbourhood of one point to another: consider homomorphism of form $\varphi(v) = v - x + y$. This proves that $V$ is $T_1$. Now, by using this property for every point $y \in Y$ we can find an open set $W_y$ such that $y \in W_y, 0\not\in W_y$. Then, just take $U = \bigcup_{y \in Y} W_y$, and this works. Mar 12, 2021 at 22:01

Let $$W$$ be a balanced neighborhood of $$0$$. Since $$+$$ is continuous, there are balanced neighborhood of $$0$$, say $$U$$ and $$V$$, such that $$U + V \subset W$$. Let $$p$$ be a cluster point of $$U$$. The $$V + p$$ is a neighborhood of $$p$$ so it contains a point of $$U$$, say $$u$$. Thus $$u = v + p$$ where $$v \in V$$. Thus $$p = u - v \in U + V$$ since $$V$$ is balanced.

Thus $$\overline{U} \subset U + V$$. This says that the collection of closed neighborhood of $$0$$ are a neighborhood base at $$0$$, so the space is regular.

• Don't see what s wrong with answer. Or have I not been complete enough with details? Jun 9, 2019 at 20:57