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-- "Every topological group is completely regular."

Teach me if I'm wrong, but I think this only holds for Hausdorff topological groups, maybe even only for locally compact groups. Since the wikipedia page for topological groups doesn't require hausdorff, this example should be removed (just in the case I'm right, for sure) --Roman3 (talk)09:47, 12 July 2010 (UTC)[reply]

The article is correct: every topological group is completely regular (with the Wikipedia conventions that neither "topological group" nor "completely regular" implies Hausdorff). This follows from the existence of auniformity on every topological group, because a topological space is uniformizable if and only if it is completely regular. --Zundark (talk)10:43, 12 July 2010 (UTC)[reply]

I corrected from

X is a completely regular space if

given any closed set F and

any point x that does not belong to F,

then

for every y in F there is a continuous function f from X to the real line R

such that f(x) is 0 and f(y) is 1.

In other terms, this condition says that x and F can be separated by a continuous function.

(This was introduced inhttp://en.wikipedia.org/w/index.php?title=Tychonoff_space&diff=next&oldid=463867677with sayingclearer if the quantifiers are in the beginning. Yes, clearer but wrong in this case.)To

X is a completely regular space if

given any closed set F and

any point x that does not belong to F,

then

there is a continuous function f from X to the real line R

such that f(x) is 0 and,for every y in F, f(y) is 1.

In other terms, this condition says that x and F can be separated by a continuous function.

Carefully with correcting from ambiguous to unambiguous - do your hit the proper case?Best regards90.180.192.165 (talk)21:10, 24 February 2012 (UTC)[reply]

The paragraph explaining the history of the concept in the lead is taken straight from Narici & Beckenstein. Now Narici & Beckenstein is a very fine book for functional analysis, but they are not historians. And in their historical commentaries they have the annoying habit of sometimes twisting the truth to get a better story. In this particular case, the notion of commpletely regular space wasnot introduced by Tychonoff. In fact, if you look at the 1930 paper from Tychonoff, Tychonoff himself mentions in a footnote that the notion was introduced in 1925 by Urysohn.

Please do not blindly reference N&B for historical stuff. They are not reliable for that.

Also, there is absolutely no need to spell out all the various transliterations of the name Tychonoff in this article. That belongs perfectly in the linked articleAndrey Nikolayevich Tychonoff, but not here.PatrickR2 (talk)04:41, 22 October 2023 (UTC)[reply]

Could you be so bold, as to edit the article, and provide the appropriate lede? Maybe add some short history section describing how things got to here?67.198.37.16 (talk)23:13, 19 November 2023 (UTC)[reply]

I don't want to edit anything before the editor who added that paragraph is made aware of this. He unfortunately has the habit of quoting Narici at every opportunity, even when it's not warranted. It would be good if he realizes he is being excessive in this. After that, we can remove most of that paragraph.

@Mgkrupa: do you care to comment?PatrickR2 (talk)06:16, 21 November 2023 (UTC)[reply]

I didn't realize that N&B was not reliable for history. Now that I know, I'll stop using them for that. Thanks for informing me. Feel free to remove that paragraph.Mgkrupa20:58, 21 November 2023 (UTC)[reply]

@Mgkrupa: Thanks for your understanding. I'll modify the paragraph.PatrickR2 (talk)03:58, 24 November 2023 (UTC)[reply]

An example of a space that is aregular space but is not completely regular would be nice to have. I'm not seeing one, just right now.67.198.37.16 (talk)23:11, 19 November 2023 (UTC)[reply]

Oh why, look! Theregular space article says:

An example of a regular space that is not completely regular is theTychonoff corkscrew.

Red link, but OK.67.198.37.16 (talk)18:31, 26 November 2023 (UTC)[reply]

TheTietze extension theorem can be applied tonormal spaces to find a continuous real-valued function that separates two closed subsets. Apparently, this theorem won't work forregular spaces, because, if it did, then every regular space would automatically be completely regular. So why does this theorem break down for regular spaces? What is the insight, intuition for this?67.198.37.16 (talk)23:24, 19 November 2023 (UTC)[reply]

The lede states that there are completely regular spaces that are not Hausdorff; and the first talk topic above suggests that this is often the case for topological groups. Can explicit examples be given? For example, by naming some topological group that is not Hausdorff?67.198.37.16 (talk)23:36, 19 November 2023 (UTC)[reply]

For a trivial example, a space with the indiscrete topology is completely regular and non-Hausdorff. Any group becomes a topological group when given the indiscrete topology. That's completely regular and non-Hausdorff.

Seehttps://topology.pi-base.org/spaces?q=completely%20regular%2B~t2%2BHas%20a%20group%20topology for other examples.

Feel free to add an example to the Examples section.

Also FYI the articletopological group already mentions something about that in the section "Quotients and normal subgroups". (see condition for when a quotient group with quotient topology is Hausdorff) (please don't repeat it in this article)PatrickR2 (talk)05:48, 21 November 2023 (UTC)[reply]

Thank you! I'll look.67.198.37.16 (talk)08:23, 26 November 2023 (UTC)[reply]

Hmm. Well, that was completely underwhelming. Thetrivial topology is, well, trivial. Pi-base listsodd-even topology anddeleted integer topology anddouble-pointed reals. Thetopological group article says less: all that it says is that when a group has a normal subgroupand the subgroup is closed (or the closure is taken) then the quotient space is Hausdorff, and "that's why we only study those." Of course, something similar would be true if we just forget about the group structure entirely. Andle voila,quotient topology says this:

If the quotient map isopen, then${\displaystyle X/\sim }$ is aHausdorff space if and only if ~ is a closed subset of theproduct space${\displaystyle X\times X.}$

but sadly, it does not state when a quotient space might be regular. Conversely, going back to groups, thetopological group#Separation properties section rules out non-Hausdorff counterexamples quickly:

As a uniform space, every commutative topological group iscompletely regular. Consequently, for a multiplicative topological groupG with identity element 1, the following are equivalent:

• G is a T₀-space (Kolmogorov);
• G is a T₂-space (Hausdorff);
• G is a T_31⁄2 (Tychonoff);
• { 1 } is closed inG;

I don't know how to make a topological group where{ 1 } isn't closed.

The articleuniform space completely dashes our hopes. Near the bottom, it says

For every topological group${\displaystyle G}$ and its subgroup${\displaystyle H\subseteq G}$ the set of leftcosets${\displaystyle G/H}$ is a uniform space. (...stuff about how to make G/H uniform...) The corresponding induced topology on${\displaystyle G/H}$ is equal to thequotient topology defined by the natural map${\displaystyle g\to G/H.}$

If I combine this with the earlier statements fromuniform space, to wit:

For a uniformizable space${\displaystyle X}$ the following are equivalent:

• ${\displaystyle X}$ is aKolmogorov space
• ${\displaystyle X}$ is aHausdorff space
• ${\displaystyle X}$ is aTychonoff space

From the above, I conclude that the only way for G/H to not be Hausdorff is for it to not be${\displaystyle T_0}$. If there is a needle to be threaded here, it's a bit too fine for me.

So I'm not seeing anything in particular that leads to something that would be completely regular, but isn't Hausdorff. So, the most interesting counter-example so far is thedeleted integer topology which I guess is safe to add to this article. I was hoping for something more ...arcane, sophisticated. Oh well.67.198.37.16 (talk)18:11, 26 November 2023 (UTC)[reply]

I see that you added a few examples to the Examples section. But please do not introduce them if you are not sure. A few comments: (1) every topological group is completely regular, even the non-commutative groups. (2) Don't introduce a link to Tychonoff corkscrew, which does not even exist in Wikipedia; and would need a reference... (3) There is no such thing as "the partition topology"; it should be "a partition topology". And the sentence as written is not even correct. I am tempted to revert all your changes, unless you want to discuss here first.PatrickR2 (talk)03:23, 28 November 2023 (UTC)[reply]

FYI for another category of examples, anypseudometrizable space is completely regular, but need not be Hausdorff in general. For example, see the function spacesLp space#Lp spaces and Lebesgue integrals defined over a suitably defined domain. These are even examples of topological groups (actually topological vector spaces, defined by a seminorm), hence completely regular, but not Hausdorff. The closure${\displaystyle H}$ of the zero function consists of the measurable functions that are zero almost everywhere. So that could be the kind of${\displaystyle G/H}$ example you were looking for.PatrickR2 (talk)03:39, 28 November 2023 (UTC)[reply]

FYI, I'm asking about this, because in dynamical systems and chaos and fractals, etc. one has some pretty wacky, singular behavior, and I'm trying to figure out how some of the singular craziness there is can be understood with conventional topological axioms. The time evolution operator in dynamical systems forms atopological monoid, not a group, and so perhaps some aspects of uniformizability are evaded??? Buttopological monoid is a stub. Meanwhile,Spectral theory of ordinary differential equations never uses the word topology, except to say the strong operator topology is used.67.198.37.16 (talk)18:38, 26 November 2023 (UTC)[reply]

It is mentioned thatT_π space is an alternative notation for Tychonoff spaces. This notation does not appear in any of the standard references. So at least it is not a notation in general use even if some author in the past may have used it once. Does anyone have a reference for this?PatrickR2 (talk)20:51, 3 December 2023 (UTC)[reply]

I'm pretty sure Császár uses this in his book General Topology, but I haven't got a copy to check at the moment. --Zundark (talk)14:08, 4 December 2023 (UTC)[reply]

The article states "every topological group is completely regular". I had modified this to read "every commutative topological group is completely regular", but this was reverted, with a statement that "... even the non-commutative groups [are completely regular]" (See above.) Is there a reference for this? Perhaps this should be obvious? It's just not obvious to me. I don't have any particular intuition for this. Is there some way to think about this that would reveal the correctness of this statement?67.198.37.16 (talk)00:50, 1 February 2024 (UTC)[reply]

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