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So while reading One Point Compactification this came to my mind.

Suppose there is a locally compact Hausdorff space $(X,\tau)$ then obviously it has a one point compactification,say, $Y$.Now, give the same space X a topology $\delta$ that is finer than $\tau$. Will $(X,\delta)$ have a one point compactification as well and if yes, will that be the same $Y\ ?$

To find the answer,if we could show whether or not $(X,\delta)$ is locally compact Hausdorff then our answer would be obtained.

Now,for being finer $\delta$ is obviously Hausdorff too. But what about local compactness?If it were only compact, then I know a finer topology may not retain compactness but not sure what happens when dealing with local compactness.How does Local Compactness depend on the fineness of a topology ?

user118494
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2 Answers2

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A topology finer than a locally compact topology may not be locally compact.

Consider the Sorgenfrey line, that is the real numbers $\mathbb{R}$ with the topology generated by the family of half-open intervals $[a,b)$ for $a < b$. This topology is finer than the usual (metric/order) topology on $\mathbb{R}$ (if $a < b$, then $(a,b) = \bigcup_{n \in \mathbb{N}} [a+\frac{b-a}{n} , b)$). However the Sorgenfrey line is not locally compact. The answers to this question characterise the compact subsets of the Sorgenfrey line, and in particular they are countable, which implies that they have empty interior. (In terms of the basic open sets, note that for any $a<b$ we have that $[a,b) = \bigcup_{n \in \mathbb{N}} [ a , b-\frac{b-a}{n} )$, and this open cover has no finite subcover.)

Taumatawhakatangihangakoauauot
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  • Sorry , I don't get it. If the compact subsets have non empty interior then that interior is an open set inside the compact set. And to be locally compact, a point $x$ needs to have a compact set that contains an open nbd of $x$. So how does non empty interior give a contradiction ? – user118494 May 09 '17 at 07:22
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    @user118494 he meant to say "empty interior". It's clear that countable sets cannot contain any set of the form $[a,b)$, which are uncountable. – Henno Brandsma May 09 '17 at 07:25
  • @user118494 Sorry, that was an error. I've fixed it in the answer. – Taumatawhakatangihangakoauauot May 09 '17 at 08:20
  • The fact that $[a,b)$ is not compact by itself doesn't show the Sorgenfrey line is not locally compact. The characterisation of compact subsets does. – Henno Brandsma May 09 '17 at 09:13
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In fact my (now deleted) post was wrong (not the initial fact) but the proof afterwards had a gaping hole.

Every space $(X, \tau)$ has a locally compact Hausdorff refinement: the discrete topology on $X$, which which almost always be strictly finer than $\tau$.

The reals in the usual topology and the Sorgenfrey topology are a classical example where a finer topology of a locally compact Hausdorff space need not be locally compact Hausdorff. But it could have been (there are no real objections).

Maybe there is a notion of "maximally LCH spaces" (which I define now, not sure if someone considered them before): $(X,\tau)$ is maximally LCH (locally compact Hausdorff) iff when $(X, \delta)$ is LCH and $\tau \subseteq \delta$, then $\delta$ is the discrete topology on $X$. (there are well-studied classes of spaces called minimal Hausdorff and maximally compact). Are there maxiamlly LCH spaces beyond the trivial (discrete) ones? Can they be characterised?

Henno Brandsma
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