In the exercise 10 of the previous section we have mentioned that a topological group must be regular but there is an example of a topological group that is not normal. So, the only question is whether it has to be completely regular or not. Here we show that the answer is positive. Recall (Supplementary exercises, Chapter 2) that there is a self-homeomorphism that maps any element to $e$ . Therefore, we only need to show that we can separate $e$ from any closed set that does not contain it by a continuous function. Also, for any neighborhood $U$ of $e$ there is a symmetric neighborhood $V$ of $e$ such that $V\cdot V\subseteq U$ . So, for every dyadic rational in $[0,1]$ define $U_p$ by the recursive procedure given in the exercise. Note that the product of any two open sets is open (as the union of open sets of the form a point $\times$ an open set), and that the product of two neighborhoods of $e$ is a neighborhood of $e$ .Let $U(0)=\{e\}$ (it is not open but just for convenience to prove the relations below, after that we assume $U(0)=\emptyset$ ). Then $U(0)\subseteq U(p)$ for all $p>0$ ($U(0)\subseteq V_n$ for all $n$ , and therefore, $U(0)\subseteq V_n\times U$ for any neighborhood $U$ of $e$ ). Now, $U(0/2^n)$ , $U(1/2^n)$ , $U(2/2^n)$ , $U(3/2^n)$ , $U(4/2^n), . . .$ = $U(0/2^{n-1})$ , $V_n\cdot U(0/2^{n-1})$ , $U(1/2^{n-1})$ , $V_n\cdot U(1/2^{n-1})$ , $U(2/2^{n-1}), . . .$ . We see immediately that $V_n\cdot U(2k/2^n)=U((2k+1)/2^n)$ . At the same time, $V_n\cdot U((2k+1)/2^n)=V_n\cdot V_n\cdot U(k/2^{n-1})\subseteq V_{n-1}\cdot U(k/2^{n-1})$ . And (by induction) the latter term is a subset of $U((k+1)/2^{n-1})=U((2k+2)/2^n)$ . Therefore, for every $n$ and $k$ , $V_{n}\subseteq U(k/2^n)\subseteq U((k+1)/2^n)$ (this obviously holds for $U(0)=\emptyset$ and outside $[0,1]$ as well).Now, given a closed set $A\subseteq G-\{0\}$ take a neighborhood $V_0=G-A$ of $e$ , define $U(p)$ as above, and $f(g)=\inf\{p=k/2^n|g\in U(p)\}$ . The only difference from the proof of the Urysohn lemma is in that we define $U(p)$ on a dense subset of set $\mathbb{Q}$ instead of the whole set $\mathbb{Q}$ , and for $p<q$ in the dense we proved there is some $n$ such that $V_n\cdot U(p)\subseteq U(q)$ instead of $\overline{U}(p)\subseteq U(q)$ . The first difference does not matter. Now, let $g\notin U(q)$ . For $v\in V_n$ , since $V_n$ is symmetric, $v\cdot g\notin U(p)$ . Therefore, $g\notin\overline{U}(p)$ and $\overline{U}(p)\subseteq U(q)$ .