Completely regularity of every topological groups. Theorem.5 P49 in T. Husain's Introduction to topological groups, says that,

Every Hausdorff topological group is completely regular.

The theorem is proved directly (Without the use of uniform spaces). I don't understand the proof of the theorem and I have a few questions, Which I can not find their answers. I am looking for a proof that is more detailed than the reference.
I have 3 problems in this theorem:


*

*Is the hausdorffness used in the proof?

*definition of $V(r)$.

*Continuity of the mapping $f$.





 A: I personally like very much the notes written by Dikran Dikranjan "Introduction to topological groups". You can download them from his web-site
http://users.dimi.uniud.it/~dikran.dikranjan/ITG.pdf
The proof you are looking for is Theorem 4.1.2 on page 23 (i do not copy his proof here as he uses some results he proved in the previous chapter).
EDIT: As noted by Thomas Winckelman in the comments, there is no "Theorem 4.1.2" in the newest version of the notes. The result I was referring to is now (April 2021) "Theorem 3.5.2" but it may change again in the future. Please note that "completely regular Hausdorff spaces" are also called "Tychonoff spaces" or "Tychonov spaces" (depending on how you transliterate from Cyrillic); Dikranjan uses this second terminology so, if you cannot find the result I mention, you can use a basic search of the word "Tychonov".
A: I'm trying to show how it can be proved in topological groups in the same way as it is proved in Uniform Spaces.
Let $(G,\mathcal T)$ be a topological group. Let
$P$ be the set of all pseudometric $d:G^2\to [0,\infty)$ with
$$(\forall r>0)(\exists  U\text{ is a neighborhood of }1)(\{ (x,y) \mid xy^{-1}\in U\}\subseteq \{(x,y)|d(x,y)\le r\})$$
For each $d\in P$, let $\mathcal T_d$ be the topology on $G$ with base
$$\{\{x| d(a,x)<r \}\mid a\in G,r>0\}$$
and $\mathcal S$ be the topology on $G$ with subbase
$$\bigcup_{d\in P} \mathcal T_d$$
If one proves $\mathcal S=\mathcal T$ then the proof in this post works. (read from "For each $d\in P$..")
So it remains to prove $\mathcal S=\mathcal T$. The part $\mathcal S\subseteq\mathcal T$ is trivial. 
The the part $\mathcal T\subseteq \mathcal S$ is very tedious. To be more specisfic, let's continue . 
Suppose $U\in \mathcal T$ is arbitrary. If $U$ is empty it is in $\mathcal S$. So let it be nonempty with $a\in U$. There's a sequence $(U_n)$ of open neighborhoods of $1$ with


*

*$U_n\subseteq Ua^{-1}$.

*$U_n=U_n^{-1}$.

*$U_{n+1}U_{n+1}\subseteq U_n$.


$\{U_nx\mid n\in \Bbb N, x\in G\}$ is a base for a topology $\mathcal T_U\subseteq \mathcal T$ on $G$. The tedious part is to prove that there's some
$d\in P$ with
$$\mathcal T_d=\mathcal T_U$$
The proof has a process similar to this. We have:
$$U\in \mathcal T_U=\mathcal T_d\subseteq \mathcal S$$
A: The proof in Hewitt,Ross, Abstract Harmonic Analysis vol.1., Structure of topological groups,... Theorem 8.4 gives more details (it is my personal impact) (almost the same proof).
