Talk:Locally compact space

Local base
I believe that the reason that some people use the definition require a local base of compact neighbourhoods is that this has been found to be a useful concept in the nonHausdorff context, whereas the definition requiring only a single compact neighbourhood has not. There is nothing in Wikipedia yet about nonHausdorff locally compact spaces in either sense, but I suggest that we set ourselves up in a position to talk about a useful concept in the future by switching the definition that Wikipedia uses. This is easy to do now; just change this and the topology glossary. &mdash; Toby Bartels, Sunday, June 9, 2002

Sure, if you think the other concept is more useful, let's change it. AxelBoldt, Sunday, June 9, 2002

Examples
I moved some of the examples over to compact space.

Is it actually true that all compact spaces are locally compact? Just checking :-)


 * See below.


 * This is the reason that I arranged the examples as I did: I wanted to highlight all the possible contrasting possibilities: spaces that were compact, spaces that were locally compact but not compact, spaces that weakly locally compact but not strongly locally compact, and spaces that weren't even weakly locally compact. Then, given the importance of the separation axioms, I wanted examples in each case of both Tychonoff and nonregular spaces. Thus:




 * I don't think it's a crime to have examples duplicated on more than one page. OTOH, you seem to have a different philosophy towards the examples entirely.


 * &mdash; Toby Bartels, Wednesday, June 26, 2002

I moved the following example:
 * Some weakly locally compact spaces that aren't locally compact: the right order topology or left order topology on any unbounded totally ordered set, in particular: the right order topology or left order topology on the set R of real numbers, useful in the study of semicontinuous functions

I believe the left order topology on R is locally compact. If x&isin;R, then ]-&infin;, x+1/n] for n&isin;N is a local neighborhood base for x consisting only of compact sets.


 * Hey, you use the backwards bracket thing too! Any chance we can get that accepted as standard practice on Wikipedia?


 * Well, your argument seems correct, but that's not what I remember from Steen &amp; Seebach; let me check again ....


 * Well, I double checked Steen &amp; Seebach, and it looks like they are using yet another definition of locally compact (their "strongly locally compact", since they use "locally compact" for the old fashioned weak sense). I had read this wrong (it is again equivalent in the Hausdorff case). This puts into doubt the claim that every compact space is locally compact, for which I was relying on Steen &amp; Seebach since I didn't see how to prove it, and also removes any indication that I have of a source that calls our local compactness "strong". (The new definition, BTW, is: every point has an open neighbourhood with compact closure.)


 * Now I want to go back and check everything over again. I apparently don't have a source that covers all 3 definitions at once. I'll report back tomorrow with what I can come up with, or you can think about it &mdash; you've got the definitions now too.


 * &mdash; Toby Bartels, Wednesday, June 26, 2002

Heh, so much for tomorrow. I'll be gone until Monday evening &mdash; then I'll work this mess out. In the meantime, visitors to the main article should be sufficiently warned. &mdash; Toby Bartels, Friday, June 28, 2002

Maybe we can avoid confusing the beginner by first explaining the standard locally compact notion for Hausdorff spaces (which is what most people need and where the definitions all agree), with all the common examples and properties, and then have a separate section where the various definitions for non-Hausdorff spaces are compared.

The reason I edited the examples was the same: I find it more important that the reader first gets a good grasp of the concept by studying simple motivating examples, rather than find themselves in an overwhelming glut of tiny distinctions that are really only important for the specialists working in the field. AxelBoldt, Friday, June 28, 2002

OK, I've written a new version now with these principles in mind. (Note that I also added the Hausdorffness condition to Wikipedia's definition of topological vector space; the simplicity that this brought to the discussion of TVSs in this article is characteristic of the reasons for that requirement in the definition.) Have at it. &mdash; Toby Bartels, Wednesday, July 3, 2002

Herrings
Some thoughts inspired by Axel's recent changes: I've always felt that the complex numbers were a red herring in functional analysis &mdash; when you get down to it, you can do most of this stuff over any commutative real C*algebra, so &iquest;what's so special about C? &mdash; which explains my focus on real-valued functions as the simplest example. But I agree that the complex-valued case is so common that it merits a mention here &mdash; especially as regards the Gelfand-Naimark theorem, the real version of which exists but is not quite as I had stated it. I do object to using notation like "C0" for the complex case as a default, but I agree that that's pretty common too. &mdash; Toby Bartels, Saturday, July 6, 2002

C* algebras
Is the classification of the commutative C* algebras really called "Gelfand-Naimark Theorem"? I seem to recall that name for the result that any C* algebra is a *-subalgebra of the algebra of linear continuous operators on some Hilbert space. AxelBoldt, Saturday, July 6, 2002

What you are citing is called "Gelfand Naimark Segal". Specifically, the Gelfand Naimark Segal construction takes a C* algebra and an element of the algebra and constructs from that a Hilbert space and a representation of the C* algebra on the Hilbert space. Your statement, the Gelfand Naimark Segal theorem, follows because the element of the algebra can be chosen so that the representation is faithful. A quick web search for "gelfand" AND "naimark" AND "theorem" brings up many references to my theorem but no references to your theorem that don't also have the name "Segal" attached. There do seem to be generalisations of the GN theorem to noncommutative C* algebras, representing the C* algebra as an algebra of functions defined on a noncommutative space (see the introduction to http://nyjm.albany.edu:8000/PacJ/1998/184-1-5.html for a survey), but this is again different from the GNS theorem &mdash; although I wouldn't be surprised if the GNS theorem couldn't be made a corollary of some of these generalisations, nor would I be surprised if the generalisations made use of the GNS construction. Still, the basic idea seems clear: a GN theorem represents the C* algebra as bounded functions on a topological space (or generalisation thereof), while a GNS theorem represents the C* algebra as bounded operators on a Hilbert space. &mdash; Toby Bartels, Saturday, July 6, 2002

Open naighbourhoods
The first paragraph has a link to "local base," which describes them as a collection of open sets. Doesn't this contradict the parenthetical sentence which asserts that the neigborhoods need not be open? Bill Kielhorn, Mar 31, 2003.


 * Yes, thanks for the catch! I'll fix it. AxelBoldt 16:53 Apr 21, 2003 (UTC)

Non-Hausdorff theory
I'm curious about the non-Hausdorff theory. I've checked about a dozen different general topology references (including Munkres, Kelley, Armstrong, Steen and Seebach, McCarty, and others) and none of them use the definition given here (in terms of a local base of compact sets). Everyone uses one of the alternate definitions: However, the article claims that the local base definition is the "best". Can anyone explain why? Or, at least give me some references that use this definition. Mostly, I am curious as to why this definition is in Wikipedia when no one seems to use it.
 * 1) Every point has a compact neighborhood
 * 2) Every point has a relatively compact neighborhood

With the current definition, it is not at all clear to me that a compact space is locally compact (this is true with either of the alternate definitions). Am I wrong, or are there any easy counterexamples? &mdash; Fropuff 19:24, 2004 Aug 12 (UTC)

Infinite dim vector spaces
K igor k: Banach-Alaoglu doesn't provide examples. You are mixing up topologies. Like in the Hilbert cube example in the text, the unit ball is not a weak neighborhood. The only way an infdim tvs can be locally compact is if it is really stupid - like having the indiscrete topology. That separated locally compact => finite dim is Chapter 1, section 13, theorem 8 of Grothendieck's TVS, or p. 65 of Edwards' Functional Analysis.--John Z 1 July 2005 05:14 (UTC)

I see, sorry for being a little obtuse. I now see the point of the Hilbert cube example. But I find that example a little misleading. If the Hilbert cube is seen as the set of sequences in [0,1], it is not even a subspace of a Hilbert space (which I think is implied to be $$l_2$$ here). Not all such sequences have finite norm. The Hilbert cube would be the unit ball in $$l_\infty$$ which is a Banach, not a Hilbert, space. -- k_igor_k (Wed Jul 6 09:21:28 EDT 2005)

Yes, you are right. If you want to use Hilbert rather than Banach spaces, the alternate definition from the Hilbert Cube article should be referred to here, sequences with nth term bounded by 1/n, to get it inside $$l_2$$, otherwise it just doesn't make sense. --John Z 7 July 2005 23:28 (UTC)

Definition
Well, I've gone and changed the definition of local compactness again. Actually, the article now mentions all three definitions up front and refuses to commit to any one. I did this because almost all references I've checked were not using our (previous) definition. The sole exception being Willard. I see no good reason to favor Willard's definition above the rest (I welcome arguments either way). If Wikipedia is to be accurate it should reflect the definitions that readers are likely to encounter in the "wild". Unfortunately, this seems to preclude choosing a favored definition here. -- Fropuff 02:49, 17 October 2006 (UTC)

expand
this page never describes, even vaguely, what a compact space is. an article should not require another article for a modicum of understanding. 66.160.66.86 (talk) 16:15, 22 January 2008 (UTC)

"one-point compactification of the rational numbers"
The article says: The one-point compactification of the rational numbers Q is compact and therefore locally compact in senses (1) and (2) but it is not locally compact in sense (3). But one-point compactification is defined only for locally compact spaces, where Q is (as mentioned some not locally compact. So, what does this sentence wants to say? -- Paul Ebermann (talk) 14:36, 8 July 2008 (UTC)
 * The construction should not require the original space to be locally compact. If Y is a topological space, define the Alexandroff extension Y* to be Y union {p}, with neighborhoods of p defined as (Y-L) union {p} where L is compact. Y* is compact, Y is contained as a subspace, Y is open in Y*, Y is dense in Y* if and only if Y is not compact, Y* is Hausdorff if and only if Y is locally compact and Hausdorff.  This is from Willard, 1970, 19A, p. 140. JackSchmidt (talk) 15:45, 8 July 2008 (UTC)
 * The construction does not, but the one-point compactification may no longer be Hausdorff. Indeed, the one-point compactification of the rationals is not Hausdorff, since any neighborhood of the point at infinity is a finite-complement neighborhood.  This should probably be addressed somewhere in the one-point compactification article, but I am reluctant to omit the requirement of local compactness from the definition.  Perhaps it should be mentioned in a separate "generalizations" section to avoid polluting the main results with pathologies.   siℓℓy rabbit  (  talk  ) 16:00, 8 July 2008 (UTC)

Examples Correct?
I am having some issues with some of the examples of "Hausdorff Spaces that are not Locally Compact" on this page, see for instance:


 * * the space Q of rational numbers (endowed with the topology from R), since its compact subsets all have empty interior and therefore are not neighborhoods;

As far as i'm aware, if some subset of $$\mathbb{Q}$$ is taken as a subset of $$\mathbb{R}$$ then sure, its interior is empty. But *within* the sub-space topology the interior of, e.g. [a,b] is just (a,b), no? A valid reason that $$\mathbb{Q}$$ is not locally compact is maybe because sequences of the form $$\left( \left[ r + \frac{\lfloor 2^n \sqrt{2} \rfloor}{2^n},r + \frac{\lfloor 2^n \sqrt{2} \rfloor + 1}{2^n} \right] \right)_{n \in \mathbb{N}, n > N}$$ for suitable rational $$r$$ and number $$N$$ form descending sequences of closed non-empty sets with empty interesection in $$\mathbb{Q}$$, and by suitable choice of $$N$$ and $$r$$ can be placed inside any neighborhood, showing that no neighborhood is compact.

... and on this example:


 * * the subspace {(0,0)} union {(x,y) : x > 0} of R2, since the origin does not have a compact neighborhood;

I'm wondering how this can be? Is there a citation (specific chapter in one of the standard books)? It seems that this is a closed subset of R and hence locally compact? Am i missing something?

199.43.12.100 (talk) 10:45, 25 April 2013 (UTC)


 * In the first example, you are right that if a and b are rationals with a < b, then the interior (in Q) of [a,b] &cap; Q is (a,b) &cap; Q. But [a,b] &cap; Q is not compact, so this doesn't contradict the assertion. As for the other example, it's not a closed subset of R2. --Zundark (talk) 11:46, 25 April 2013 (UTC)

Okay looking at both examples more closely i see that the wording of the article is correct, thanks. 199.43.13.100 (talk) 12:12, 25 April 2013 (UTC)