# Is “compact implies sequentially compact” consistent with ZF?

Over at the nForum, we've been discussing sequential compactness. The discussion led me to realise that I naively assumed that nets were simply Big Sequences, and that I could make a reasonable guess at how nets would behave by thinking of them as such.

Not so. The crucial point, that I hadn't realised, was that subnets are not sub-nets in the way that subsequences are sub-sequences.

Where this came to light was in a discussion of the relationship between compactness and sequential compactness. Compactness can be expressed as:

Every net has a convergent subnet.

Sequential compactness as:

Every sequence has a convergent subsequence.

So, in my naivety, I assumed that compactness implied sequential compactness since I could take a sequence, think of it as a net, find a convergent subnet, and - ta-da - there's my convergent subsequence. The error, as Mike Shulman pointed out, is that not every subnet of a sequence is a subsequence.

And, indeed, there is a space that is compact but not sequentially compact. Writing $I = [0,1]$ then $I^I$ is compact but not sequentially compact. In particular, it is possible to find a sequence that has no convergent subsequence (the argument is a variant of Cantor's diagonal theorem) but that has plenty of cluster points and thus plenty of convergent subnets.

But the compactness of $I^I$ seems to require a Big Axiom (not quite the axiom of choice, or so I'm led to believe since $I$ is Hausdorff, but almost). I say "seems to" since I'm not an expert and there may be a way to prove that this specific space, $I^I$, is compact with only the basic axioms of ZF.

That's basically my question, except that I'm a topologist so I'm more interested in the implications for topological stuff than in the exact relationship between the Axiom of Choice and Tychanoff's theorem (and since I can just read the nLab page to learn that!). So, without further ado, here's the question:

Is "Compactness => Sequential Compactness" consistent with ZF?

This could be answered by a topologist since all it would require to show that this isn't so would be an example of a space that was compact but not sequentially compact and such that proving that didn't require any Big Axioms.

References:
1. nLab pages: sequential compactness (has more details on the above example), nets (contains the crucial definition of a subnet), Tychonoff's theorem (contains a discussion of the axiomatic strength of this theorem)
2. nForum discussion: sequential compactness
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Another classical example of a non-sequentially compact but compact space is the Cech-Stone compactification of N. In this space the only convergent sequences are the eventually constant ones. This is a "common" way to get non-sequentially compact... – Henno Brandsma May 12 '10 at 21:45
Sort of a stupid remark, but have you flipped through Counterexamples In Topology to see if the matter is resolved there? – Pete L. Clark May 12 '10 at 23:58
Pete: I looked. The examples they give of spaces that are compact but not sequentially compact owe their compactness to Tychonoff's theorem. – Nate Eldredge May 13 '10 at 3:08
@Nate: OK, that's not too surprising, but thanks for checking. – Pete L. Clark May 13 '10 at 3:19

The sequential compactness of $[0,1]^{\omega_1}$ is undecidable in ZFC: as noted above $[0,1]^{[0,1]}$ is not, so under CH $[0,1]^{\omega_1}$is not sequentially compact; on the other hand $\mathrm{MA}+\neg\mathrm{CH}$ it is sequentially compact. Thus the question is still open.

$\mathrm{MA}$ implies that any product of fewer than continuum many sequentially compact spaces is sequentially compact. In the case of $\aleph_1$ many and when $\mathrm{MA}+\neg\mathrm{CH}$ is assumd you follow the proof for products with countably many factors and produce, given a sequence $(x_n)$ in the product, infinite subsets $A_\alpha$ of $\mathbb{N}$ such that $(x_n)$ restricted to $A_\alpha$ converges on the first $\alpha$ coordinates and such that $A_\alpha\setminus A_\beta$ is finite whenever $\beta<\alpha$. $\mathrm{MA}+\neg\mathrm{CH}$ now implies there is an infinite set $A$ such that $A\setminus A_\alpha$ is finite for all $\alpha$. Then $(x_n)$ restricted to $A$ converges in the full product.

A very nice introduction is still Mary Ellen Rudin's article in the Handbook of Mathematical Logic.

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As I said in the question, I'm a topologist not a set theorist, so please pardon the basic clarification request. I've heard of "CH", but not "MA". What's the full name? – Loop Space Jun 23 '10 at 12:52
Martin's Axiom (en.wikipedia.org/wiki/Martin's_axiom) is a combinatorial axiom that often provides an alternative to CH. There are numerous examples where a statement implied by CH is refuted under $MA+\neg CH$. – Joel David Hamkins Jun 23 '10 at 13:19
Thanks to both of you (Joel and KP) for the clarification. – Loop Space Jun 23 '10 at 20:56

If you let $X=\prod_{\mathbb{R}}[0,1]$. By Thychonoff's Theorem this is compact. But one can construct a nice sequence by diagonalization such that it has no converging subsequence: let such a subsequence consist of a bunch of $0$'s and $1$'s and it does not converge in $[0,1]$. This is basically your example.

Now if we don't want AC then just let $X=\prod_{\omega_1}[0,1]$: no need for AC here, this is compact in ZF. Indeed $\omega_1$ is already well ordered. This space is not sequentially compact.

So compactness does not imply sequentially compactness in ZF.

Now what I wrote might be complete nonsense I am just a beginner at that stuff but eh! why not try to write an answer!

Edit: any stuff that is going to look like the Stone-Cèch or involve ultrafilters is going to need some choice, so maybe we need an example that is not related to the structure of the Stone-Cèch

Edit #2: hold on, what if we take an arbitrary product of Tychonoff Planks. So basically it looks like that $X=\prod_{A}[0,\omega_1]$X$[0,\omega$]. This is compact. Is this sequentially compact? This is does not look like first countable.

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Regarding your second example, I don't think that ZF proves Tychonoff for products over a well-ordered index set. Isn't this a weak choice principle? – Joel David Hamkins May 12 '10 at 23:13
When you carry out the proof, at successor steps you can take the set of $x\in X_{\beta}$ such that $x_{\beta}\cup {x}$ is not a good point i.e $U_{x_0}$ can't be covered finitely by stuff of the cover of the product}}(when you take your cover in the product)This set is closed in [0,1] because its complement is open You can take x($\beta$) to be the least like that in the usual order of [0,1]. – Carlo Von Schnitzel May 12 '10 at 23:31
I don't quite follow what you say. What I am thinking is this: the usual proof that Tychonoff implies AC does not involving changing the index set, so Tychonoff for products indexed by I implies choice for families indexed by I. Thus, if ZF proved Tychonoff for countable products, it would imply countable choice, which it doesn't. – Joel David Hamkins May 13 '10 at 0:13
This may be a really stupid question, but: Without AC, do we even know that $\omega_1$ exists? – Harald Hanche-Olsen May 13 '10 at 0:33
Harold, yes, you don't need AC for that. $\omega_1$ is the set of countable ordinals. One can prove this exists just in ZF. For example, using the Powerset and Comprehension axioms, one forms the set of all well-order relations on $\omega$, and each is order isomorphic to a unique countable ordinal (by Replacement), so the set of such ordinals exists by Replacment. No AC needed. – Joel David Hamkins May 13 '10 at 0:46

This Problem is already solved. See Horst Herrlich: "The Axiom of Choice" Springer

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How about a page number of the book? Or is the whole book on this one question? – Gerald Edgar Feb 19 '13 at 13:44