Thanks, Guest!

Have added this to the entry, now here and here.

(Also to *compactly generated topological space*: here and here)

You can drop the Hausdorff assumption: Escardó, Lawson, and Simpson (doi:10.1016/j.topol.2004.02.011) state that a space is a *k*-space iff it is a colimit in Top of compact Hausdorff spaces (lemma 3.2.iv, p. 110 = p. 7) iff it is a quotient of a locally compact Hausdorff space (definition 3.3 and corollary 3.4.iii, p. 111 = p. 8).

That would imply that a Hausdorff k-space is a colimit of compact Hausdorff spaces, no? Locally compact Hausdorff spaces are colimits of compact Hausdorff spaces, as noted on locally compact topological space just above where you inserted the statement there.

]]>this entry used to vaguely indicate, without proof or refernce that, that Hausdorff k-spaces are the colimits of diagrams of locally compact Hausdorff spaces.

I have added to that (here) a precise statement with a pointer to a proof.

]]>added the statement (here) that: Every locally compact Hausdorff space is compactly generated and weakly Hausdorff.

]]>Thanks, Jürgen. Yes, I agree this entry is probably due for a clean-up.

]]>So, my reason for the preference of 2.1 is that it separates the issues `locally compact' and `regular'.

Jürgen Voigt, Dresden ]]>

You may be thinking that there are (convex) bounded open neighborhoods of the origin in your Frechet space example, whose closure would also bounded. But that’s just not so: a TVS with such neighborhoods is *normable*, which is not the case with your example. See theorems 7 and 8 here.

Isn’t the Heine-Borel property just that closed and bounded implies compact? This doesn’t imply local compactness.

In any case, I am quite sure that locally compact Hausdorff TVS are finite-dimensional. For a source of this claim outside the nLab, see for instance here.

]]>Not every topological vector space that is Hausdorff and locally compact is finite dimensional. An example is the the Frechet space of Holomorphic functions in an open neighborhood. This space has the Heine-Borel property. See also Montel Spaces.

Anonymous

]]>You’re welcome! Nice one!

]]>Thanks!! I have fixed it now.

]]>I.e. you need to replace $K$ on the left of the product by an open neighbourhood of $x$ which it contains.

]]>In the proof of continuity of the exponentiation map at Introduction to Topology – 1 that you linked to, it is claimed that $K \times V^{K}$ is open. This does not parse: you would need $K$ to be both compact and open, and local compactness does not give that.

]]>Not sure what you are pointing me to. I say “there is a compact subset which is still a neighbourhood” and “neighbourhood” implies “contains an open neighbourhood”. But I may be missing what you are trying to tell me.

]]>By the way, Urs, I think the same issue affects your proof at Introduction To Topology, you may wish to fix it there.

As I think I mentioned once before, I am entirely content with replacing the proof I wrote at compact-open topology with your re-write :-).

]]>I have fixed the proof of 3.1 at compact-open topology now. Note that when we say that every neighbourhood has a compact neighbourhood in Definition 2.1 at locally compact topological space, it is vital in the second use of neighbourhood that we mean that it contains an open neighbourhood of $x$, not just that it is any old set which contains $x$. Otherwise the proof of 3.1 does not go through (one cannot take $U''$ to be $V$, which was what was written before I edited just now).

]]>Hi Urs, unless I’m missing something, it’s not quite right yet. I’ll try to fix it now.

]]>Thanks to you two for checking!!

It seems you are leaving the task of editing the proof in the entry to me. I did so now: removed the extra clause about forming the closure of $V$ in def. 2..5 here, so that now $V$ is already a compact sub-neighbourhood. Then I replaced all following occurences of $\overline{V}$ with $V$.

Notice that the writeup and style of the proof is very much Richard’s. I tend to feel I get a bit lost in the notation there. I feel more comfortable with my own wording of Richard’s proof, here. That’s just to say: better check my edits (as long as nobody else edits the entry, these edits are highlighted in green and red here).

]]>Actually, now I agree as well. All you need to do is replace the statement by an interpolation of type $U \subseteq K \subseteq V$, and I think it’s fine. So very good.

]]>I believe that I agree with Urs, that the proof of 3.1 at compact-open topology goes through perfectly well using 2.1 at locally compact topological space instead, and this is the only place that local compactness is used. So I am happy to use 2.1 instead of 2.2 for that (but it would be nice to make a remark saying that one can use 2.2).

PS - I corrected a typo and fixed a reference at compact-open topology.

]]>But do we need later on that the compact neighbourhood in that step 2 is a closure?

]]>Okay, so the immediate question that comes up is how you get step 2 in the proof of 3.1. It doesn’t look true for an infinite set with the cofinite topology, which is locally compact in the sense of definition 2.1.

]]>Urs, I’d like to go through your proof carefully. Off-hand I’m a little bit skeptical if the assertion is that the exponential topology matches the compact-open topology in quite that generality.

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