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Let $X$ denote the product of countably many copies of $[0,1]$. Let $X_1$ denote the set $X$ with the box topology and $X_2$ denote the set $X$ equipped with the product topology.Then which ones are correct?

  1. $X_1$ is compact and separable
  2. $X_2$ is compact and separable
  3. $X_1 $ and $X_2 $ are both compact
  4. Neither $X_1$ nor $X_2$ is separable

I know a topological space is compact if every open cover has a finite subcover and separable means the space has a countable dense subset.I know $[0,1]$ is compact and finite product of compact spaces is compact, but for infinite case I don't know anything. Please help. Thank you.

BLAZE
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user274289
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    This is not going to be easy if you don't allow for some slightly more advanced topological machinery, you might want to research the standard Tychonoff theorem, then e.g. have a look at this http://math.stackexchange.com/questions/89790/tychonoff-theorem-in-the-box-topology and this http://math.stackexchange.com/questions/97413/on-the-product-of-separable-spaces – Damian Reding Sep 26 '15 at 14:45

1 Answers1

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  • Every product of compact spaces is compact. That's standard Tychonoff theorem.
  • Every product of at most continuum many separable spaces is separable. That's special case of standard Hewitt–Marczewski–Pondiczery theorem.
  • No infinite box product of nondegenerate Hausdorff spaces is compact. That follows from the fact that the standard product is Hausdorff compact and the box product topology is strictly finer, so it cannot be compact (standard result that every continuous map from a compact space to a Hausdorff space is closed).
  • No infinite box product of nondegenerate Hausdorff spaces is separable. If $X_i$, $i ∈ I$, are the spaces and $U_{i, 0}, U_{i, 1}$ are nonempty disjoint open in $X_i$, then $\{∏_{i ∈ I} U_{i, f(i)}: f ∈ \{0, 1\}^I\}$ is a family of at least continuum nonempty disjoint sets open in box topology.
user87690
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