Is $\prod_{n=1}^{\infty} A_{n}$ with product topology, where $A_{n}$=${\{0,1\}}$ has discrete topology for , $n = 1,2,3,\cdots.$ a compact set? How to show it?

My approach: To show that a set is compact, we need to show that every open cover has finite subcover. Now with respect to discrete topology open sets are either singletons or whole space. For Example $A_1 \times A_2\times A_3\times \cdots $ can be covered by $(\{0\}\cup \{1\}) \times (\{0\}\cup \{1\})\times \cdots $ which has finite subcover for each of the $A_i$ but there are countably infinite cartesian product, so what can we conclude from this? However each $A_i$ is itself finite and hence compact w.r.t discrete topology. Is my approach right.

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    $\begingroup$ From Tychonoff theorem, the product of an arbitrary number of compact spaces is compact in the product topology. Every $A_n$ is compact. That's it! $\endgroup$ – Kolmin Oct 13 '16 at 17:23
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    $\begingroup$ @Kolmin the OP is apparently asking for a direct proof of Tychonoff theorem in this special case, where you only have countably many spaces. $\endgroup$ – J.-E. Pin Oct 13 '16 at 18:28
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    $\begingroup$ @J.-E.Pin: Feel free to correct (or add to) my feedback to OP. :). To OP: I am not sure what is your approach, in the sense that it seems to me that you are restating the definition of compactness, and then you are describing the situation, but you did not mention (or I cannot see) what is your attack strategy, Said so, if I should have to prove it without Tychonoff, I would proceed with sequential compactness, which is equivalent to compactness for metrizable spaces (and this one is obviously metrizable via discrete metric). $\endgroup$ – Kolmin Oct 13 '16 at 21:06

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