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Is $[0,1]^\omega$ with product topology a compact subspace of $\mathbb{R}^{\omega}$, where $\mathbb{R}^{\omega}$ denote the space of countably many products of $\mathbb{R}$. Is the subspace locally compact?

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    $\begingroup$ What does $\Bbb{R}^\omega$ have to do with anything? Being compact/locally compact is an intrinsic property of a space. It doesn't matter where it's embedded. $\endgroup$ – Chris Eagle Jun 15 '13 at 21:04
  • $\begingroup$ @ChrisEagle Hi. Why is it not valid ? I remember in Munkres topology the following defintion was stated: Let $X$ be a topological space $A\subseteq X$ is said to be a compact subspace of $X$ iff for every set of open subsets of $X$ that cover $A$ have a finite subset of open sets that cover $A$ $\endgroup$ – Amr Jun 15 '13 at 21:21
  • $\begingroup$ @Chris Eagle Thus I think the terminology is OK. Am I right ? $\endgroup$ – Amr Jun 15 '13 at 21:22
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    $\begingroup$ @StefanH.: $\mathbb{R}^\omega$ is not locally compact. $\endgroup$ – Nate Eldredge Jun 15 '13 at 21:30
  • $\begingroup$ @NateEldredge: You are right, thank you! I was too over-hasty and didn't think this through. But it it true for finite products. $\endgroup$ – Stefan Hamcke Jun 15 '13 at 21:38
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By Tychonoff's theorem, the product of any family of compact spaces is compact, so $[0,1]^\omega$ is compact. Compactness is a property that a space possesses or not, independent of the surrounding space, so it is a compact subspace of $\Bbb R^\omega$, given that the product topology on $[0,1]^\omega$ is the same as the topology as a subspace. That seems like a trivial fact, and a proof that a product of subspaces is a subspace of the product is indeed easy, using the universal property of the initial topology.

$[0,1]^\omega$ is also locally compact, as is every product of compact locally compact spaces. Here locally compact means that every point has a local base of compacts sets.

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    $\begingroup$ Note that if you work with the definition "locally compact iff every point has a compact neighbourhood", then compact trivially implies locally compact, so the last statement doesn't have much content in that case. $\endgroup$ – kahen Jun 15 '13 at 23:33
  • $\begingroup$ That's right. I will include the definition in my answer. $\endgroup$ – Stefan Hamcke Jun 15 '13 at 23:46
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    $\begingroup$ Not every product of locally compact spaces is locally compact. The product $X$ of locally compact spaces $X_i$ is locally compact iff all but finitely many of them are compact (which is the case here, as all are compact). $\endgroup$ – Henno Brandsma Jun 16 '13 at 6:02
  • $\begingroup$ @HennoBrandsma: That's why I wrote "compact locally compact spaces". $\endgroup$ – Stefan Hamcke Jun 16 '13 at 13:58

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