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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:

  1. Is the hausdorffness used in the proof?

  2. definition of $V(r)$.

  3. Continuity of the mapping $f$.


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Theorem 3, S20.

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  • $\begingroup$ @MinimusHeximus: Yeah, you are right (Every topological group can be viewed as a uniform space). But my problem is that the hausdorffness used in the proof or not? I guess when he says "Since $e\in V_r$ for all $r\in D$" he used hausdorffness,But I'm not sure. $\endgroup$
    – TXC
    Jul 13 '13 at 3:12
  • $\begingroup$ I haven't looked through the entire proof to verify whether Hausdorffness is used in it, but I can say something about whether the theorem itself requires Hausdorffness, which is: it all depends on what you mean by completely regular! $\endgroup$ Dec 6 '16 at 7:44
  • $\begingroup$ The broadest definition of completely regular is that a space is completely regular iff, for each point x and closed set C with x not in C, there is a continuous real-valued function f such that f(x) = 0 and f[C] = {1}. That's certainly what most of this proof is about. But another definition states that a space is completely regular iff [that stuff above about continuous functions] and the space is Hausdorff! With that definition, the group must be Hausdorff just because a completely regular space must be Hausdorff. $\endgroup$ Dec 6 '16 at 7:46
  • $\begingroup$ In any case, the well-known theorem is that every topological group is completely regular in the broader sense, and every Hausdorff group is completely regular in the stricter sense. If this proof is sufficiently general, then its use or nonuse of Hausdorffness will reflect this. $\endgroup$ Dec 6 '16 at 7:48
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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".

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  • $\begingroup$ It appears that a "Theorem 4.1.2" doesn't exist, and that nothing immediately applicable is presented on page 23. Also, passing "completely regular" into ctrl+F turns up zero results. Would it be reasonable to ask if you elaborated somewhat? $\endgroup$ Apr 9 at 2:25
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    $\begingroup$ Well, some 8 years have passed since that answer was written. Since then, prof. Dikranjan has changed and improved his notes several times. In their actual version I think that the theorem I was referring to is the new Theorem 3.5.2. You cannot find anything looking for "completely regular" just because "completely regular Hausdorff spaces" are also called "Tychonoff spaces" and this is the name under which they are studied in those notes. $\endgroup$
    – Simone
    Apr 9 at 11:32
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    $\begingroup$ And indeed I did not know that! Thanks for the update. (boy, it's strangely easy to forget sometimes that the internet is 8 years old) $\endgroup$ Apr 9 at 15:58
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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

  1. $U_n\subseteq Ua^{-1}$.
  2. $U_n=U_n^{-1}$.
  3. $U_{n+1}U_{n+1}\subseteq U_n$.

$\{U_nx\mid n\in \Bbb N, x\in G\}$ 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: $$U\in \mathcal T_U=\mathcal T_d\subseteq \mathcal S$$

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  • $\begingroup$ I want this answer be deleted. $\endgroup$
    – user79193
    Sep 30 '13 at 14:48
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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).

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