Since $X$ is a compact metric space, it is sequentially compact so every sequence in $X$ has a subsequence which converges. Now, I want to show that $X^\mathbb{N}$ is sequentially compact, but I'm getting confused because we must show that the sequence of a sequence has a convergent subsequence. I know that we must use the fact that $X$ is sequentially compact. Thanks!
If $X$ is a compact metric space, then the set of all sequences in $X$, $X^\mathbb{N}$ is sequentially compact
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1Well, $X^N$ is in fact *compact* (for the reasosonable topology...) – 2011-03-17
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0This is a direct result of Tychonoff's theorem. – 2011-03-17
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0Yes I need to prove that it is compact and I thought it would be easier to prove that it is sequentially compact. The space I am trying to prove is sequentially compact (or compact since we proved it is a metric space) is the set of all sequences in X. I don't think it is the product of space like R^n. Any further explanantions would be appreciated. It is supposed to involve some diagonalization argument – 2011-03-17
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0@Mariano Suárez-Alvarez, @Alex: Please see my answer for a reply to your comments. – 2011-03-17
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0I just reread your comment here and noticed I'd previously overlooked the sentence "I don't think it is the product of space like R^n". The set of all sequences in a space is (isomorphic to) the product of countably many copies of the space. Just like an element of an $n$-fold product is an $n$-tuple, an element of an $\mathbb{N}$-fold product is an $\mathbb{N}$-tuple, i.e. a sequence. – 2011-03-17
1 Answers
The answers in the comments are somewhat misleading in two respects.
For one, they don't mention that they use the facts that a) the space of sequences in a metric space is metrizable and b) compactness and sequential compactness are equivalent in metrizable spaces. The desired result can only be proved from Tychonoff's theorem using these facts; the product of an uncountable number of compact metric spaces is in general not metrizable and not sequentially compact.
But more importantly, they obscure the set-theoretic premises required for the desired result. Tychonoff's theorem in full generality is equivalent to the axiom of choice. We're only dealing with a countable product here, but I believe Tychonoff's theorem for countable products still requires the boolean prime ideal theorem and the axiom of countable choice (see this article, p. 2). By contrast, the diagonal argument the OP was looking for requires no choice (not even countable choice):
Given a sequence $s$ of sequences in $X$, denote by $s_i(n)$ the $n$-th element of the $i$-th sequence of $s$. (If you tend to get as confused about such things as I do, it helps to keep in mind the concrete example where $s_i(n)$ is the $n$-th digit in the binary representation of $i$.) Consider the sequence $s_i(1)$ in $X$. Since $X$ is a compact metric space and hence sequentially compact, this sequence contains a convergent subsequence $s_{i_1(k)}(1)$. Denote its limit by $a (1)$. Then consider the sequence $s_{i_1(k)}(2)$ (a subsequence of the sequence $s_i(2)$ in $X$). This sequence, too, has a convergent subsequence $s_{i_2(k)}(2)$ with limit $a (2)$. We can continue this construction for all $n$. Now consider $s_{i_l(l)}$, which is a subsequence of the sequence $s$ of sequences. We can show that it converges against the sequence $a$ of the limits $a (n)$.
An element of the basis of the product topology containing $a$ is a product of factors, only a finite number of which differ from $X$. For each $n$, apart from the first $n$ elements the sequence $s_{i_l(l)}(n)$ is a subsequence of $s_{i_n(l)}(n)$, and thus it is a convergent subsequence of $s_i(n)$ with limit $a (n)$, which will eventually end up in the $n$-th factor of the basis element. Since there are finitely many of these factors, eventually all these convergent subsequences will have ended up in their corresponding factors, so that the entire sequence eventually ends up in the given basis element.
[Edit in response to comments by Nate and Theo:]
Here's a construction that picks a particular convergent subsequence of a sequence having at least one convergent subsequence in a metric space without making arbitrary choices (in case Nate and Theo are right that that requires the axiom of choice):
For a given sequence $b_n$, consider the (non-empty) set of all convergent subsequences for which $\lvert b_n - b \rvert < 2^{-n}$ for all $n$, where $b$ is the limit of the subsequence, and take the least of these subsequences in lexicographical order. We can't just take the lexicographically least convergent subsequence because there isn't one, but we can explicitly construct the lexicographically least convergent subsequence that fulfills the given condition: For each element, just pick the earliest element in the sequence such that the resulting partial subsequence is a prefix of at least one subsequence that fulfills the condition. The result is a Cauchy sequence, and hence, since $X$ is compact, converges. (The same construction doesn't work without the additional condition, since the resulting subsequence would just be the sequence itself.)
So it appears that we can avoid choice. However, that would seem to allow us to reason as follows: The fact that all factors in the product are $X$ was not used. Thus, the result generalizes to arbitrary countable products of compact metric spaces. A countable product of metric spaces is metrizable, and in a metrizable space compactness and sequential compactness are equivalent; thus we've proved that the countable product of compact metric spaces is compact. But now we can plug that into the proof of the equivalence of the axiom of choice and Tychonoff's theorem (linked to above). Since to any compact metric space we can add an isolated point and still have a compact metric space (by limiting the metric to $1$ and choosing a uniform distance to the new point $>1$), this seems to prove the possibility of choice from a countable product of compact metric spaces. This is not the full axiom of countable choice, but it does allow considerable choices that I would have expected to be independent of ZF. Is this true, or is there another fallacy somewhere?
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1Ah, but you *are* using (countable) choice. The sequence $s_{i_l(k)}(l)$ doesn't just contain *one* convergent subsequence, it contains a lot of them. So you have to choose one. There are infinitely many choices to be made... In fact, if you do your argument with pseudometric spaces instead, it is known to be equivalent to countable choice (this is Form 8AK in Howard and Rubin's web site; see http://consequences.emich.edu/conseq.htm). – 2011-03-17
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0@Nate: Thanks for this, I was about to make the exact same point. – 2011-03-17
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0@Nate Eldredge @Theo Buehler: Thanks. Three remarks: a) unless I'm mistaken that Tychonoff's theorem for countable products requires more than countable choice, my point about not using Tychonoff remains valid. b) The equivalence for pseudometric spaces that you cite is a simple modification of the equivalence proof I linked to, since the topology used there is induced by a pseudometric with distance $0$ within $A_i$ and distance $1$ to $x$, but you can't do the same thing for compact metric spaces, since these don't exist for arbitrary cardinalities, so I'm note sure the result is relevant. – 2011-03-17
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0@Nate Eldredge @Theo Buehler: c) There may well be a fundamental flaw in my understanding of the axiom of choice here. I thought that I only need to invoke choice if I claim that there's a set containing the chosen representatives. But I'm not making such a claim here; I'm just using a single subsequence and winnowing it down further and further, without ever constructing a set of choices. If you're saying that this, too, requires the axiom of choice, that makes its negation seem even more counterintuitive than it did before :-) – 2011-03-17
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0@Nate Eldredge: BTW, there seems to be a mistake in that Form 8 AK statement, "Countable products of pseudometric spaces are compact" -- I think it should read "*compact* pseudometric spaces"? – 2011-03-17
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0@Nate Eldredge @Theo Buehler: I updated the answer to avoid arbitrary choices. I'd be interested to hear whether the construction is OK now, and also what you make of the reasoning in the last paragraph. – 2011-03-17
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0@joriki: to be honest, I don't understand how not mentioning two absolutely standard facts is *misleading*. Moreover, it appears that *you* are interested in knowing the "set-theoretic premises required for the desired result" but I see no such interest in the OP: as far as what he wrote, he is just asking a question about good ol' metric spaces. – 2011-03-17
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0@Mariano Suárez-Alvarez: I'm sorry if "misleading" was too strong. I did say "somewhat misleading", though :-). Do I understand correctly that you take the axiom of choice for granted when you deal with "good ol' metric spaces"? That's certainly one valid view on the axiom of choice, but we both don't know what the OP's view on it is. Also, if you assumed she or he knew such "absolutely standard facts", what was the point of pointing out an even more standard fact? – 2011-03-17
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1@Joriki: if someone has trouble showing that a countable product of compact metric spaces is compact, it is a safe bet that he does not care much about how much choice is needed in order to prove the result---and in fact a discussion about that can most probably confuse more than help. Specially when there is nothing in his question to indicate otherwise! (Usually people that *do* care about choice are very vocal about it :) ) – 2011-03-17
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0@Mariano Suárez-Alvarez: It's also a safe bet that you have more experience with people learning topology than I do, so you may well be right. The way I learned topology, Tychonoff's theorem was inseparable from the axiom of choice, but that may have been atypical. – 2011-03-17
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0@Mariano Suárez-Alvarez: Having said that, I do think that even if the OP (or someone else reading the question and answers) wasn't currently interested in the axiom of choice, there's some value in avoiding to use it when it's not necessary -- at some point she or he might get interested in it, and then it will be unnecessarily hard to disentangle what makes essential use of it and what doesn't. – 2011-03-17
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0@joriki, @Mariano: I'm with Mariano and Nate here. However, I do agree that the diagonal argument is the way to go about this question (after all, the passage to universal nets/ultrafilters in the usual proofs of Tychonov are also a sort of diagonalization). I simply wouldn't try to sell the diagonal argument by making a point about choice because this tends to make things rather more complicated and intransparent than more lucid, as the present case shows yet again. That's all. (actually, I voted the answer up when I made the comment) – 2011-03-17
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1@Theo Buehler: Thanks for your comment (and the vote :-). But after you said that my claim that I hadn't used choice was flawed, I'd appreciate your opinion as to whether the flaw is now fixed. (Since there seems to be some level of consensus here that these issues shouldn't be discussed in response to this question, I'm thinking of opening a new question to clarify these issues.) – 2011-03-17
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0I'm not yet entirely convinced by the argument and I'm too tired now to think it through in full detail. I'll have a look at it again tomorrow or on Saturday (I'll be quite busy tomorrow). I think your idea is very nice and maybe opening a new question is a better way to go about it because I'm definitely no expert on avoiding choice (I'm more of a geometer/analyst by training). – 2011-03-17
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0@Theo Buehler: Thanks! I'll open a new question then (my first one :-). Sleep well! – 2011-03-18
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0Thanks, you too (it's getting late in Berlin too, isn't it? :)) By the way, you can call me Theo if you want, I'm quite a bit younger than you. – 2011-03-18
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0@Theo: I see -- I wasn't actually using your last name out of formality but because I thought that I have to use your full user name if I want you to get notified of the message :-) Yes, it's getting late... – 2011-03-18
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0I think I read somewhere that the first three letters suffice. Here you can find some explanations and further links: http://meta.math.stackexchange.com/questions/1694/ Good night you too! Looking forward to your question. – 2011-03-18