Show that every topological group is $T_3$

Question

I know that it is sufficient to show that for one point ($e$) and any neighbourhood $U$ of $e$, we have a neighbouhood $V$ with $\bar{V} \subseteq U$. Since $x \to x^{-1}$ is continuous, it follows from continuity of multiplication that we have neighbourhoods $V_e$ and $V_e^{-1}$ of $e$ such that $V_e \circ V_e^{-1} \subseteq U$. However, I don't know how to show that this implies that $\bar{V_e} \subseteq U$. Every source I have found just states that the last inclusion is true, but I don't understand why?

Answer 1

(I’m assuming that $e$ is the identity element.)

Let $x\in\operatorname{cl}V_e$; $xV_e$ is an open nbhd of $x$, so $xV_e\cap V_e\ne\varnothing$. Thus, there are $v_0,v_1\in V_e$ such that $v_1=xv_0$, and hence $x=v_1v_0^{-1}\in V_eV_e^{-1}\subseteq U$.

Answer 2

Generalizing this a bit. In a topological group $G$ we have for all sets $A\subseteq G$ the recipe $$ \overline{A}=\bigcap_{U\in D} AU, $$ where $D$ is the family of symmetric open neighborhoods of $e$, and $AU=\{au\mid u\in U,a\in A\}$. This is because all the neighborhoods of a point $x\in G$ contain a neighborhood of the form $Ux$ for some $U\in D$. Symmetricity of $U$ means that $Ux$ has a non-empty intersection with $A$ iff $x\in AU$.

In particular $\overline{A}\subseteq AU$ for all such $U$.

Apply this to $A=V_e$ and $U=V_e^{-1}$.

Answer 3

The following are historical comments on the classical proof given above by user "Brian M. Scott".

Any topological group is regular but need not be Hausdorff; the statement of the question assumes the separation axiom $T_3$ but only requires regular (certain modern textbooks unfortunately add Hausdorff to the definition of regular). An example of a non-Hausdorff topological group is the group $C_2$ of order two equipped with the indiscrete topology; it is regular but not $T_3$.

Also, this same proof appears in Lev Pontryagin's 1938 book, "Непрерывные Группы" ("Continuous Groups"), as Замечание $\S$16:F. In the second edition (1954), in Пример 32, Pontryagin credits the discovery of the proof to Andrey Kolmogorov and states that it was discovered by him shortly after the notion of topological group was formulated.

In fact, Pontryagin then credits complete regularity (if also Hausdorff, this would be $T_{3.5}$, also known as Tikhonov) to unnamed others of the same era. There exist topological groups $G$ that are not normal. However, the Birkhoff--Kakutani theorem (1936, independently discovered by both) states that any topological group $G$ admits an invariant metric if and only if $G$ is $T_0$ and first-countable. (Note $T_0$ here is equivalent to $T_2$, since any translation in $G$ is a homeomorphism.)