# Tychonov's theorem in countable spaces!?

Give a direct proof of Tychonov's theorem: If $(X_n, d_n)$ is compact, then $\left(\prod_{n\geq1} X_n, d\right)$ is compact.

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Is this a question, or a direct command? –  Asaf Karagila Nov 18 '12 at 16:27
I changed ($\prod_{n\ge1}X_n$,$d$) to $\left(\prod_{n\ge1}X_n,d\right)$. That's the right way to use TeX. –  Michael Hardy Nov 18 '12 at 17:58

Let $d_n'=1/(1+d_n)$. The metrics $d$ and $d'$ generate the same topology. Define a metric on the product space by $d^*\big((x_n),(y_n)\big)=\sum_n 1/2^n d_n'(x_n,y_n)$. One can chech that $d^*$ indeed metrizes the product topology. So the countable product is metrizable and compactness and sequential compactness coincide.
We show that the product is sequentially compact. Let $(x_n)$ be a sequence of points, i.e. sequences, in the product space. Let $(y_n^1)$ be a subsequence that converges in the first coordinate, $(y_n^2)$ a further subsubsequence that converges in the second coordinate, and, by construction also in the first coordinate. Continuing this way, we get a sequence $\big((y_n^1),(y_n^2),(y_n^3),\ldots\big)$ of subsequences. We construct now a convergent subsequence of all these subsequences as $(y_1^1,y_2^2,y_3^3,\ldots)$. This sequence converges in every coordinate of the product and therefore in the product topology.