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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.

Martin Sleziak
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Patrick Da Silva
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    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. – Martin Jun 23 '13 at 13:16
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    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. – David Mitra Jun 23 '13 at 13:17
  • @David Mitra : So you are saying it's a non-trivial result, if I understood things well? – Patrick Da Silva Jun 23 '13 at 13:36
  • @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... – Patrick Da Silva Jun 23 '13 at 13:41
  • No. I'm not familiar enough with the material to give an opinion. I was merely providing a reference. – David Mitra Jun 23 '13 at 13:41
  • @David Mitra : Okay, well if I can't figure out I'll go take a look at that book at my university's library tomorrow! – Patrick Da Silva Jun 23 '13 at 13:42
  • The key fact is that every neighborhood $U$ of the identity contains a symmetric neighborhood $W$ such that $W^2 \subseteq U$ (this basically expresses continuity of multiplication and inversion at the identity). If $F$ is a closed set not containing the identity, choose a decreasing sequence $V_0 \supseteq V_1 \supseteq \cdots$ of symmetric neighborhoods of the identity such that $V_{k+1}^2 \subseteq V_{k}$ and $F \cap V_{0} = \emptyset$. Then the onion shell argument will produce a continuous function vanishing at the identity and equal to one outside $V_0$, which is what you want. – Martin Jun 23 '13 at 13:51
  • @Martin : Yes, I understand the onion argument, the reason why I fail to see the argument in the general case is that I don't understand how I can get that decreasing sequence to be strictly decreasing... I guess that's necessary so that the 'onion shells' are distinct? – Patrick Da Silva Jun 23 '13 at 14:13
  • You need to prove first that a ($T_0$) topological group is regular. Then apply regularity to ${e}$ and $G\setminus V_{k}$ and get your strictly decreasing sequence. – Martin Jun 23 '13 at 14:24
  • @Martin : But not all topological groups are $T_0$, aren't they? Is there a way to reduce the proof from general topological groups to the Hausdorff case by quotienting by the closure of ${e}$? – Patrick Da Silva Jun 23 '13 at 14:34
  • You don't really need $T_0$ (and no, topological groups need not be $T_0$, take a seminormed space which is not a normed space). Here's the argument to show that topological groups are $T_3$ from the fact above: Suppose $V$ is open and does not contain $e$. Choose a symmetric and open $W \ni e$ such that $W^2 \subseteq V$. For all $x \in \overline{W}$ we have $W \cap xW \neq \emptyset$ so that there are $w_1,w_2 \in W$ such that $w_1 = xw_2$. Then $x = w_1 w_{2}^{-1} \in WW^{-1} = WW \subseteq U$. Therefore $e \in W \subseteq \overline{W} \subseteq U$. – Martin Jun 23 '13 at 14:39
  • @Martin : I know not all topological groups are $T_0$, it's just that you seemed to assume it so I was kind of disturbed. – Patrick Da Silva Jun 23 '13 at 14:53
  • @Martin : Oh, I didn't expect regularity to be that simple to prove... (I've proved all the propositions in my link so I knew that for an open set $U$ I had the existence of $W$ with $W \subseteq \overline{W} \subseteq WW \subseteq U$, I just didn't think of letting $U = G \backslash \mathcal F$ where $\mathcal F$ would be closed.) – Patrick Da Silva Jun 23 '13 at 14:57
  • @Martin : I gave it a little thought, I don't think I need the chain to be strictly decreasing! If I take a finite group with the discrete topology, it will be completely regular by taking an injective function $f$ whose image is made of isolated points, and that function will work for every $x \in G$ and $\mathcal F \subseteq G$ closed, but of course my decreasing sequence of open sets needs to be constant up to some point. So I think I'll be fine just taking the decreasing sequence of symmetric open sets... – Patrick Da Silva Jun 23 '13 at 17:18
  • Yes. The sequence will be strictly decreasing unless you can take some $V_k$ to be clopen -- and if this can happen the entire discussion is pretty much moot :-) – Martin Jun 23 '13 at 19:15
  • Although topological groups need not be $T_0$, they might as well be for many purposes, including this problem. The point is that the closure $C$ of (the singleton of) the identity is a normal subgroup. Take the quotient group by $C$ and give it the quotient topology. Unless I'm having a really bad day, the resulting quotient is a $T_0$ topological group. Furthermore, any closed or open subset of the original group is a union of cosets of $C$and thus corresponds to a closed or open subset of the quotient group. – Andreas Blass Jun 24 '13 at 01:28
  • @AndreasBlass : Yes, and I know that, it is out of the question. But thank you for noticing :) the fact you just gave is one of the reasons why I am interested in topological groups ; any topological group $G$ is such that $G/C$ becomes a $T_{3 \frac 12}$ space, and the only step in the proof I was missing was the complete regularity. – Patrick Da Silva Jun 24 '13 at 01:46
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    @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. – M.Sina Jun 24 '13 at 14:28
  • @M.Sina : Yes, but the point is that every topological group (not necessarily Hausdorff) is also completely regular. Hence my question. But I'll give it a read to see if it inspires me any techniques of proof. Thanks! EDIT: I thought it was an online file. I won't bother. – Patrick Da Silva Jun 24 '13 at 14:44
  • @PatrickDaSilva: I'm not sure but I think that hausdorffness is not used. You can download it here:-). – M.Sina Jun 24 '13 at 15:12
  • @M.Sina : Hausdorffness is used when they assume that ${e}$ has a countable neighborhood base ; look at the theorem right above the proof. – Patrick Da Silva Jun 24 '13 at 16:51

3 Answers3

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Have a look at this:

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

Problem 2 is what you want.

Vishal Gupta
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  • A bit late, I have about 2 or 3 proofs by now... but hey, thanks! – Patrick Da Silva Aug 27 '13 at 18:28
  • Can you share your proofs with us? – Vishal Gupta Aug 28 '13 at 01:42
  • 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. – Patrick Da Silva Aug 28 '13 at 04:20
  • I thought you had some "other" proofs apart from ones given here. – Vishal Gupta Aug 28 '13 at 07:10
  • 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 : http://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. – Patrick Da Silva Aug 28 '13 at 07:32
  • I just thought you had some other proofs in addition to the one I linked or the one others gave here. So I suggested you to post it. I do not wish you to explain it. – Vishal Gupta Aug 28 '13 at 13:30
  • @Vishal : Sure. Good idea, but unfortunately I don't have different proofs. :P – Patrick Da Silva Aug 28 '13 at 20:10
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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

Community
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  • 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. – Patrick Da Silva Jun 30 '13 at 22:10
  • 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. – Patrick Da Silva Jul 01 '13 at 03:28
  • 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. – Patrick Da Silva Jul 01 '13 at 14:21
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To go with user79193's proof that uniformizable spaces are completely regular, here's a proof that topological groups are uniformizable.

For each open neighborhood $U$ of the identity $e$, let $D_U=\{(x,y):x y^{-1}\in U\}$. We'll show $\{D_U:U\text{ open}\}$ is a base for a uniformity inducing the topology.

First we show it's a base.

  1. $\Delta\subseteq D_U$ since $xx^{-1}=e\in U$.

  2. $D_U\cap D_V=\{(x,y):x y^{-1}\in U\cap V\}=D_{U\cap V}$.

  3. Given an open neighborhood $U$ of $e$, let $V$ be an open neighborhood of $e$ such that $V^2\subseteq U$. Then $D_V^2=\{(x,z):\exists y(xy^{-1},yz^{-1}\in V)\}\subseteq\{(x,z):\exists y(xy^{-1}yz^{-1}\in V^2)\}\subseteq\{(x,z):xz^{-1}\in U\}$ $=D_U$.

  4. $D_U^{-1}=\{(x,y):y x^{-1}\in U\}=\{(x,y):xy^{-1}yx^{-1}yx^{-1}\in xy^{-1}U yx^{-1}\}$ $=\{(x,y):xy^{-1}\in xy^{-1}U yx^{-1}\}=D_{xy^{-1}U yx^{-1}}$, noting that $xy^{-1}Uyx^{-1}$ is an open neighborhood of $e$ since $xy^{-1}eyx^{-1}=e$.

To see that it induces the correct topology, let $U$ be an open neighborhood of $x$: it follows that $xU^{-1}$ is an open neighborhood of $e$ and $D_{xU^{-1}}[x]=\{y:xy^{-1}\in xU^{-1}\}=U$.

Steven Clontz
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