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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?

- Axiom of Choice and Cartesian Products
- The Continuum Hypothesis & The Axiom of Choice
- Why is the axiom of choice separated from the other axioms?
- Zorn's Lemma and Injective Modules
- Do proper dense subgroups of the real numbers have uncountable index
- Dense subset of the plane that intersects every rational line at precisely one point?

Thanks for all the input in advance.

- Advantage of accepting non-measurable sets
- 3 Statements of axiom of choice are equivalent
- Constructiveness of Proof of Gödel's Completeness Theorem
- Infinite Set is Disjoint Union of Two Infinite Sets
- Axiom of Choice and Right Inverse
- Compact Metric Spaces and Separability of $C(X,\mathbb{R})$
- Zorn's Lemma $\equiv$ Axiom of Choice
- Choice function for a collection of nonempty subsets of $\{0,1\}^\omega$
- Prob. 9, Sec. 4.3 in Kreyszig's Functional Analysis Book: Proof of the Hahn Banach Theorem without Zorn's Lemma
- Is there an explicit isomorphism between $L^\infty$ and $\ell^\infty$?

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.

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.

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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