Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
Very nice! thanks! –  Johan Nov 18 '12 at 18:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.