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I'm working on a theorem on compactness for uniform spaces. Here are the definitions I'm using:

$X$ is compact if every open cover of $X$ reduces to a finite subcover.

$X$ is filter-compact if for every proper filter $\mathcal{F}$ on $X$, there exists a proper filter on $X$ that is a convergent refinement of $\mathcal{F}$.

$X$ is complete if every proper Cauchy filter on $X$ is convergent.

$X$ is totally bounded if for every entourage $U$ of $X$, there exists a cover of $X$ by a finite collection $\{V_i\}_{i=1}^n$ of subsets of $X$ such that we have $V_i \times V_i \subseteq U$ for each $i$.

$X$ is limit point compact if every infinite subset of $X$ has a limit point.

Here's what I believe I've managed to do so far:

For a uniform space $X$, the following are equivalent:

  • $X$ is compact;
  • $X$ is filter-compact;
  • $X$ is complete and totally bounded.

It'd be very nice if limit point compactness could be added to the above list.

I've seen this post, which suggests that (open set) compactness, sequential compactness, and limit point compactness aren't equivalent. I've been hoping that replacing sequential compactness by filter-compactness is enough to make everything work, but I've made no progress.

Does anyone know if it's doable? If yes, I'd really appreciate a hint. If not, a counterexample would be great. Thanks!

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  • $\begingroup$ Filter compactness is just a equivalent formulation of compactness in all topological spaces. $\endgroup$ – Henno Brandsma Jan 7 at 22:52
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There are limit point compact spaces that are not compact, even uniformisable spaces, e.g. take $X=\omega_1$ (the first uncountable ordinal in the order topology (that Munkres calls $W$ IIRC)). $X$ is also sequentially compact.

I don’t see how adding an equivalent formulation (filter compactness), which is equivalent in all topological spaces to standard (open cover ) compactness, is going to change anything. $X$ is not compact so it’s not filter compact, period. If $S$ is any uniform space, being limit point compact as a topological space is not going to make it compact.

Both the equivalence of compact and filter compact for all topological spaces and the equivalence of compactness with “complete and totally bounded” in all uniform spaces are classic and well-known. What the OP states is just a combination of these two facts.

Sequential compactness “orthogonal” to compactness: one need not imply the other in general spaces. “Filter compactness” is just another way to say compactness, don’t let an analogy with sequences mislead you here. “Net compactness “ (every net has a convergent subnet”) is also just a reformulation of compactness for all spaces. Sequences are “special”.

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  • $\begingroup$ Thanks, I realise now that I confused a lot of things. It's a lot clearer now! By the way I have yet another question on uniform spaces that I will be posting shortly; I've seen your name appear on another website that answered this question too, so I hope you could take a look at it for me $\endgroup$ – jessica Jan 10 at 7:10
  • $\begingroup$ @jessica I already did. $\endgroup$ – Henno Brandsma Jan 10 at 13:50

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