# Heine-Borel theorem for $\mathbb{R}^\omega$?

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. – Jonas Meyer Nov 28 '11 at 6:27
@JonasMeyer:(Silly comment: since the space is a topological vector space, there is a definition of bounded which does not depend on a metric :) ) – Mariano Suárez-Alvarez Nov 28 '11 at 6:38
@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?) – Mariano Suárez-Alvarez Nov 28 '11 at 6:39
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... – Mariano Suárez-Alvarez Nov 28 '11 at 6:52
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. – Mariano Suárez-Alvarez Nov 28 '11 at 21:41

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! – Feri Nov 29 '11 at 3:52

[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.

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