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I was answering a question recently that dealt with compactness in general topological spaces, and how compactness fails to be equivalent with sequential compactness unlike in metric spaces.

The only counter-examples that occurred in my mind required heavy use of axiom of choice: well-ordering and Tychonoff's theorem.

Can someone produce counter-examples of compactness not being equivalent with sequential compactness without the use axiom of choice? Or is it even possible?

Thanks for all the input in advance.

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The ordinals are still well-ordered even without the axiom of choice, and they are still well-founded. This means that as a topological space $\alpha+1$ is still compact, and if $\alpha$ is a limit ordinal then $\alpha$ is still not compact.

Sequential compactness talks about countable subsets, so if we take $\omega_1$ it is closed under countable limits and therefore sequentially compact, but as a limit ordinal it is not compact. Note that for that to be true we need to assume a tiny bit of choice - namely $\omega_1$ is not a countable union of countable ordinals.

Without the axiom of choice we can have strange and interesting counterexamples, though. One of them being an infinite Dedekind-finite set of real numbers. Such set cannot be closed in the real numbers so it cannot be compact. However every sequence has a convergent subsequence because every sequence has only finitely many distinct elements.

There is a section in Herrlich's The Axiom of Choice in which he discusses how compactness behaves without the axiom of choice. One interesting example is that in ZFC compactness is equivalent to ultrafilter compactness, that is every ultrafilter converges.

However consider a model in which every ultrafilter over $\mathbb N$ is principal. In such model the natural numbers with the discrete topology are ultrafilter compact since every ultrafilter contains a singleton. However it is clear that the singletons form an open cover with no finite subcover.

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  • $\begingroup$ @Nate: Thanks. I just gave a seminar about this a few days ago so it's all so fresh in my mind! $\endgroup$ – Asaf Karagila May 16 '12 at 19:32
  • $\begingroup$ Wy is $\omega_1$ sequentially compact without choice? $\endgroup$ – Chris Eagle May 16 '12 at 19:37
  • $\begingroup$ @Chris: Hmmm... I suppose that you are correct. We need to assume that it is regular. I will add that. $\endgroup$ – Asaf Karagila May 16 '12 at 19:39
  • $\begingroup$ Thanks for the excellent reply. I was aware of compactness being equiv with ultrafilter compactness in ZFC, but I will look up Herrlich's material for sure. The infinite Dedekind-finite set sounds very interesting. What do you think would be the best source to look it up from? $\endgroup$ – T. Eskin May 16 '12 at 21:17
  • $\begingroup$ @Thomas: Herrlich has quite a lot on that in his book. I think there is no "canonical" place to learn about D-finite sets, which have a rich back story and quite the theory (which completely disappear in the presence of even countable choice). I suggest you to start with Herrlich and see how it goes from there. $\endgroup$ – Asaf Karagila May 16 '12 at 22:55
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The first uncountable ordinal $\omega_1$ is sequentially compact in the order topology, since every sequence in $\omega_1$ is bounded below $\omega_1$ and there will be a first ordinal $\alpha$ containing infinitely many members of the sequence, which will hence be a limit of a subsequence of the sequence. But $\omega_1$ is not compact, since $\omega_1$ is the union of the open initial segments, and this cover has no finite subcover.

Similarly, the long line is sequentially compact but not compact.

The proof that $\omega_1$ exists does not require any use of the axiom of choice---it is completely constructive.

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  • $\begingroup$ Proof that $\omega_1$ exists needs no choice, but how can you prove that $\omega_1$ isn't the limit of a sequence of countable ordinals without choice? $\endgroup$ – Chris Eagle May 16 '12 at 19:37
  • $\begingroup$ Oops! You're right! One needs some choice (countable choice suffices) to know that $\omega_1$ is regular. $\endgroup$ – JDH May 16 '12 at 19:48
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I answered question as you did, and while reading about sequential compactness, I found that this matter has been discussed on Mathoverflow.

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