Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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
share|improve this question
2  
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
2  
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

3 Answers 3

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.

share|improve this answer
    
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
1  
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

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

share|improve this answer
5  
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

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.

share|improve this answer
    
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

Your Answer

 
discard

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.