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I have a question: does the Heine-Borel theorem hold for the space $\mathbb{R}^\omega$ (where $\mathbb{R}^\omega$ is the space of countable sequences of real numbers with the product topology). That is, prove that a subspace of $\mathbb{R}^\omega$ is compact if and only if it is the product of closed and bounded subspaces of $\mathbb{R}$ - or provide a counterexample.

I think it does not hold. But I can't come up with a counterexample! Could anyone please help me with this? Thank you in advance.

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    Specifying the topology does not specify the metric, or whether or not the metric is bounded. Every metric space is homeomorphic to a metric space with bounded metric, and for a noncompact bounded metric space, it is clear that not every closed and bounded subspace is compact. So please specify which metric you are using.2011-11-28
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    Dear Jonas, thank you for your comment and pointing this out. Yes, a metric should be specified. But in this question, a metric has not been specified. I guess the Euclidean metric is assumed. Or maybe the square metric! Does the answer depend on it? Thanks.2011-11-28
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    @JonasMeyer:(Silly comment: since the space is a topological vector space, there *is* a definition of bounded which does not depend on a metric :) )2011-11-28
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    @Farokh, your question says «prove that a subspace of $\mathbb R^\omega$is compact iff it is the product of closed and bounded subspaces of $\mathbb R^\omega$» and that does not make sense. In particular, that statement has no resemblance whatsoever with the usual Heine-Borel theorem for $\mathbb R^n$! (The Euclidean metric does not make sense for $\mathbb R^\omega$, by the way... and what is the square metric?)2011-11-28
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    @MarianoSuárez-Alvarez: Ah, good point! Farokh: Well, it turns out to not matter in *this* case, only because it is not true regardless of the metric.2011-11-28
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    @Mariano, I agree with you. I think it implies to give a counterexample. If my comments are not very accurate, sorry about that, because I'm a physics student!2011-11-28
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    What do you agree with? That the statement you want to prove does not make sense? One cannot find counterexamples for statements that do not make sense...2011-11-28
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    Oh, I see!!! This is a question from my homework!2011-11-28
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    @Farokh: You can edit your question to fix the statement Mariano refers to to be what you intended. Just use the "edit" link at the bottom left of the post.2011-11-28
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    @Mariano, Munkres defines the square metric for R^n as follows: Rho(x,y)=max{|x1-y1|, ..., |xn-yn|}.2011-11-28
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    @Farokh: That definition only makes sense when $n$ is a positive integer, and similarly for the Euclidean metric. An analogue of the "square" metric for sequences would be the sup metric, which can be defined on the subspace of bounded sequences. Similarly for the Euclidean metric, which could be defined on the subspace of square-summable sequences, but does not make sense on $\mathbb R^\omega$ as Mariano said.2011-11-28
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    Hey guys, Tychonoff theorem states that an arbitrary product of compact spaces is compact in the product topology. So, if we take [a,b], which is compact in R, then [a,b]^omega is compact in R^omega by Tychonoff theorem. Right? But I think [a,b]^omega is not bounded? Is it right or not? So, can it be a counterexample?2011-11-28
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    @Jonas, OK, I see! I should really distinguish R^n and R^omega.2011-11-28
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    @Farokh: If $(X,d)$ is a metric space, then to show that $X$ does not have the Heine-Borel property by giving a counterexample would mean to find a subspace $A\subseteq X$ such that $A$ is closed in $X$, $A$ is bounded in the metric $d$, and $A$ is not compact. You will not find (and do not want to find) an example of a compact subspace that is unbounded. It is always true that a compact subspace of a metric space is closed and bounded. Note, again, that in order to give an explicit counterexample, it must be specified what the metric is. Again, "bounded" depends on the metric.2011-11-28
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    OK, I just talked to the instructor. And as you guys pointed out, there was some typo in the question. I fixed it: That is, prove that a subspace of $\mathbb{R}^\omega$ is compact if and only if it is the product of closed and bounded subspaces of $\mathbb{R}$ (NOT $\mathbb{R}^\omega$) - or provide a counterexample.2011-11-28
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    So, if we now assume the Euclidean metric, can we give a counterexample?2011-11-28
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    The "corrected" statement is also false (it is not true even if we replace $\omega$ by $2$...) and, again, has no similarity to the usual Heine-Borel theorem.2011-11-28
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    It would not hurt you much if you googled for the statement of the Heine-Borel theorem, really!2011-11-28
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    OK, I think now I realized what you mean. From Munkres, "Heine-Borel theorem: A subspace A of $\mathbb{R}^n$ is compact iff it is closed and is bounded in the Euclidean metric d or the square metric Rho". It does NOT say that it is the product of the closed and bounded subspaces of $\mathbb{R}$. Is this what you are saying?2011-11-28
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    If it is the case, I (in fact the instructor) should NOT mention the Heine-Borel theorem. She should have just asked this question: "Prove that a subspace of Rω is compact if and only if it is the product of closed and bounded subspaces of R - or provide a counterexample". Which is a false statement as Jim Conant gave a counter example below.2011-11-28
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    @Mariano, If my comments above are right, please let me know so that I can fix this question (that is, remove the Heine-Borel theorem from question itself).2011-11-28
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    Thank you all for clarifying this question for me!2011-11-29

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The statement that a subspace of $\mathbb R^\omega$ is compact if and only if it is the product of closed and bounded subspaces of $\mathbb R$ is false even for $\mathbb R^2$. Take the "plus sign" subset $(\{0\}\times [-1,1])\cup ([-1,1]\times\{0\})$. It is compact but not a product of subsets of $\mathbb R$. This can be easily generalized to $\mathbb R^\omega$ via the inclusion $\mathbb R^2\hookrightarrow\mathbb R^\omega$ given by $(x,y)\mapsto (x,y,0,0,0,\ldots)$.

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    I think $S^1$ in $\mathbb{R}^2$ also does the job!2011-11-29
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[Edit: This answer does not answer the question. I tried an answer before the question was clarified, and it turns out to be an answer to the wrong question.]

A metric space with the Heine-Borel property (that every closed and bounded subspace is compact) must be locally compact and $\sigma$-compact, because closed balls are compact in such a space. Because $\mathbb R^\omega$ is neither locally compact nor $\sigma$-compact, it does not have the Heine-Borel property under any compatible metric.

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One thing you can say is that a subset of $\mathbb{R}^\omega$ is compact iff it is closed and contained in a product of bounded sets. I'll leave the proof as an exercise.

More generally, let $X_i$ be any family of Hausdorff spaces (and assume the axiom of choice). Then a subset $A$ of $X = \prod_i X_i$ is compact iff $A$ is closed and contained in a product of compact sets.