Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
+1 for the infinite Dedekind-finite set. – Nate Eldredge May 16 '12 at 19:30
@Nate: Thanks. I just gave a seminar about this a few days ago so it's all so fresh in my mind! – Asaf Karagila May 16 '12 at 19:32
Wy is $\omega_1$ sequentially compact without choice? – Chris Eagle May 16 '12 at 19:37
@Chris: Hmmm... I suppose that you are correct. We need to assume that it is regular. I will add that. – Asaf Karagila May 16 '12 at 19:39
+1. Excellent response. – Mathemagician1234 May 16 '12 at 20:23

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.

share|cite|improve this answer
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? – Chris Eagle May 16 '12 at 19:37
Oops! You're right! One needs some choice (countable choice suffices) to know that $\omega_1$ is regular. – JDH May 16 '12 at 19:48

I answered question as you did, and while reading about sequential compactness, I found that this matter has been discussed on Mathoverflow.

share|cite|improve this answer
Thanks; that link contains some nice discussions. – T. Eskin May 16 '12 at 21:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.