Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
3  
@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
3  
@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
1  
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
1  
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

3 Answers 3

up vote 2 down vote accepted

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 $(x,y)\mapsto (x,y,0,0,0,\ldots)$.

share|improve this answer
    
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.

share|improve this answer

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.

share|improve this answer

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.