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. 
 A: [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.
A: 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.
A: 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 $(x,y)\mapsto (x,y,0,0,0,\ldots)$.
