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Define $A_n=\{0,1\}$ for $n=1,2,3,\cdots$ and suppose it is given discrete topology, then will $\prod^{\infty}_{n=1} A_n$ with product topology be compact?

Since $A_n$ is compact, by Tychonoff's theorem the product will be compact. Am I correct?

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  • $\begingroup$ Yes, that's correct. $\endgroup$ – David Mitra Dec 21 '15 at 8:42
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    $\begingroup$ Tychonoff's theorem $\endgroup$ – user295959 Dec 21 '15 at 8:43
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Yes. Being finite, $A_n$ is compact, since for all subsets of $A_n$ each of its open covers is finite (and thus it is its own finite subcover).

Tychonoff's theorem assures compactness of arbitrary products of compact spaces, therefore also $\prod\limits_{n\in\mathbb{N}} A_n$ is compact.

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    $\begingroup$ Good. However, since Tychonoff's theorem depends on the axiom of choice, one may wish to give a direct proof for the space $\prod_{n-1}^\infty A_n$ by showing that it is homeomorphic to the Cantor set, a closed and bounded subset of $\mathbb R.$ $\endgroup$ – bof Dec 21 '15 at 9:45
  • $\begingroup$ Nice remark! I prefered to keep it simple for OP $\endgroup$ – Ottavio Bartenor Dec 21 '15 at 9:55
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    $\begingroup$ Discrete is superfluous. Finite is enough for compactness. $\endgroup$ – Henno Brandsma Dec 21 '15 at 10:32
  • $\begingroup$ True. Editing that $\endgroup$ – Ottavio Bartenor Dec 21 '15 at 10:33

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