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I am studying topological groups, and I have been able to do quite a lot on my own by proving the propositions in this link on my own, but when I read up wikipedia that topological groups are all completely regular, I wasn't able to either find a proof or do it myself, which got me concerned.

The question is ; can anyone confirm me that topological groups are indeed completely regular, and if so, where could I find a proof, or do you know it?

One proof I would be interested in is that accordingly to this Wikipedia page concerning uniform spaces, a topological group can be equipped with the structure of a uniform space in a canonical way, and uniform spaces are completely regular. I don't know how to prove this one either.

I would also be interested in a more direct approach if possible for the purposes of giving a talk on topological spaces ; if a direct proof is shorter than the one going through uniform spaces, it would give me more time in my talk to mention other things.

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    $\begingroup$ The essential ideas are contained in this blog post by Terry Tao on the Birkhoff-Kakutani theorem (see his Remark 3). A detailed proof that topological groups are completely regular (via uniform structures and Birkhoff-Kakutani) can be found e.g. in Hewitt-Ross, Abstract harmonic analysis, Vol. 1. Theorem 8.4. $\endgroup$ – Martin Jun 23 '13 at 13:16
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    $\begingroup$ That topological groups are uniformilizable is problem 35F in Willard's General Topology. That uniform spaces are completely regular is Theorem 38.2 of the same book. $\endgroup$ – David Mitra Jun 23 '13 at 13:17
  • $\begingroup$ @David Mitra : So you are saying it's a non-trivial result, if I understood things well? $\endgroup$ – Patrick Da Silva Jun 23 '13 at 13:36
  • $\begingroup$ @Martin : Terry Tao's post seems to be interested in spaces that are metrizable, not uniformizable, and that first countable hypothesis doesn't make me feel very comfortable. How do I pull off this 'very similar argument' he mentions if $G$ is not first countable? That seems to me to be the key in the argument, the rest is an 'onion construction' of the function $f$ which makes $G$ completely regular... $\endgroup$ – Patrick Da Silva Jun 23 '13 at 13:41
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    $\begingroup$ @PatrickDaSilva: Theorem.5 P49 in Introduction to topological groups, says that Every Hausdorff topological group is completely regular, And the theorem is proved directly. The proof given is too complicated for me and i have a hard time with the details. Hopefully be useful to you. $\endgroup$ – M.Sina Jun 24 '13 at 14:28
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Every uniform space $(X,\mathcal D)$ is completely regular.

sketch of a proof: Suppose $F$ is closed in $(X,\mathcal D)$ and $p\in F^c$. There's a pseudometric uniformity $P$ on $X$, such that: $$\mathcal D=\mathcal D_P=\bigcup_{d\in P}\mathcal D_d$$ Where $\mathcal D_d$ is the usual uniformity by the pseudometric $d:X^2\to [0,\infty)$.

For each $d\in P$, define $$f_d:X\to\Bbb R$$ $$f_d(x)=\inf_{c\in F}d(c,x)$$ and $$g_d:X\to \Bbb R$$ $$g_d(x)=d(p,x)$$ $f_d$ and $g_d$ are continuous. It's not hard to prove there's some $d_0\in P$ with $$(\forall a\in X)(f_{d_0}(a)\ne 0\text{ or } g_{d_0}(a)\ne 0)$$ This may help. Define $$h:X\to [0,1]$$ $$h(x)=\frac{g_{d_0}(x)}{g_{d_0}(x)+f_{d_0}(x)}$$ $h$ is continuous and $$h(p)=0,\quad h(F)=\{1\}$$


Edit:

linked

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  • $\begingroup$ To be honest, I don't understand much about uniform spaces ; the reason why I was interested about the proof with uniform spaces is because I've practically never used them and I thought this would be a place to start. Do you have a suggestion about somewhere to read the basics concerning uniform spaces? I can't read your proof yet, uniform spaces still mean nothing to me right now. $\endgroup$ – Patrick Da Silva Jun 30 '13 at 22:10
  • $\begingroup$ To be honest this document feels more like a reference than an explanation of the theory. I feel like I have to work everything out and it's hard to guess what you have to understand... I don't want to find my path in the forest, I want to follow the steps of those who went in before me! If you know what I mean. $\endgroup$ – Patrick Da Silva Jul 1 '13 at 3:28
  • $\begingroup$ I guess I'll have to do the same thing I did with topological groups and do all the work myself! It has been fun to do it with topological groups, I hope it'll be just as fun with uniform spaces. $\endgroup$ – Patrick Da Silva Jul 1 '13 at 14:21
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Have a look at this:

http://www.math.wm.edu/~vinroot/PadicGroups/519probset1.pdf"

Problem 2 is what you want.

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  • $\begingroup$ A bit late, I have about 2 or 3 proofs by now... but hey, thanks! $\endgroup$ – Patrick Da Silva Aug 27 '13 at 18:28
  • $\begingroup$ Can you share your proofs with us? $\endgroup$ – Vishal Gupta Aug 28 '13 at 1:42
  • $\begingroup$ Well aren't they all here? :P Why would you need that? I'm guessing you can solve the one you linked, and the second one is given in the answer by Minimus Heximus on this question. Look out for comments for different approaches too. $\endgroup$ – Patrick Da Silva Aug 28 '13 at 4:20
  • $\begingroup$ I thought you had some "other" proofs apart from ones given here. $\endgroup$ – Vishal Gupta Aug 28 '13 at 7:10
  • $\begingroup$ If you want to see a full detailed proof, the question you linked, I answered it and typed it up in a document. Feel free to look at it in my topological groups document available on academia.edu : math-berlin.academia.edu/Patrick1DaSilva At this point I'm confused if you actually know how to prove it or if you would like to see a different proof from the one you linked. $\endgroup$ – Patrick Da Silva Aug 28 '13 at 7:32

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