# Countable product of finite sets with a new metric, compact?

Suppose we have a finite set $E$. Is it true that $E^n$ is compact? The metric on $E^n$ is : $$d(\omega,\omega\prime)=\begin{cases}2^{-\inf \{ n \in \mathbf N:\omega _n \ne \omega'_n\} }&{\omega \ne \omega '}\\ 0&{\omega = \omega '} \end{cases}$$

My idea is as below:

For an arbitrary sequence, projection on the first element will give a one dimensional sequence which has a repetitive subsequence. Considering this subsequence in the main infinite dimensional sequence one can continue this process to find for each $n$, a subsequence which has fixed $n$ elements in its first $n$ places. Unfortunately I cannot convince myself this solution can be extended for the infinite case(the whole sequence). Can someone give me a reason why this is correct for the whole sequence or do I need something else rather than this reasoning?

Thank you.

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Note that $d(\omega,\omega')<2^{-n}$ iff $\omega_k=\omega_k'$ for all $k\le n$. It follows immediately that the topology generated by $d$ is the product topology on $E^{\Bbb N}$, considered as a product of discrete $|E|$-point spaces. As such it is compact by the Tikhonov's product theorem.

Your argument is doing it the hard way, but it does work. Suppose that $\sigma=\langle\omega^n:n\in\Bbb N\rangle$ is a sequence in $E^{\Bbb N}$, where $\omega^n=\langle\omega_k^n:k\in\Bbb N\rangle$. As you say, there is an $e_0\in E$ such that $$A_0=\{n\in\Bbb N:\omega_0^n=e_0\}$$ is infinite. Suppose that for some $m>0$ we have already chosen $e_k\in E$ and infinite sets $A_k\subseteq\Bbb N$ for each $k<m$ in such a way that $A_{k+1}\subseteq A_k$ for $k=0,\dots,m-1$, and $\omega_k^n=e_k$ for all $n\in A_k$. Then there is an $e_m\in E$ such that $$A_m=\{n\in A_{m-1}:\omega_m^k=e_m\}$$ is infinite, and the recursive construction goes through to yield a sequence $\omega=\langle e_n:n\in\Bbb N\rangle\in E^{\Bbb N}$. It’s straightforward to check that $\omega$ is a cluster point of $\sigma$.

(By the way, this argument is essentially that used to prove König’s lemma, of which this is an immediate consequence.)

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Thanks for your answer. I would not consider my way the hard way because Tychonoff's problem has such hard proof, and using that for this looks like to kill a fly with bazooka! But one thing I don't understand about your proof is what you mean by recursive construction are you referring to: en.wikipedia.org/wiki/… –  Naji Mar 18 '13 at 23:50
@Naji: (The Tikhonov theorem is actually pretty easy to prove if you learn about convergence in terms of filters.) A recursive construction/definition is one that proceeds in stages, at each stage constructing/defining something in terms of things that have already been defined. For example, a recursive construction/definition of the factorial function is given by $n!=n(n-1)!$ for $n>0$, and $0!=1$. Here I constructed $\omega$ recursively, one term at a time, and the definition at each stage depended on what I’d already done at earlier stages. –  Brian M. Scott Mar 18 '13 at 23:56
@Naji: I prefer the more accurate transliteration Tikhonov of the Cyrillic Тихонов. –  Brian M. Scott Mar 19 '13 at 0:02
Well I am afraid that wouldn't work! That recursive method you are referring to is for finite stages and would not help for infinite case. For any $n$ you can prove the existence of your sequence but for infinity(whole sequence) you need something like limit I believe. I read these things many years ago and I cannot recall correctly. I didn't know it has such a spelling. Thanks for letting me know! –  Naji Mar 19 '13 at 0:14
@Naji: It most certainly does work; it’s a bog-standard basic technique, though you’ll have to take a serious set theory course if you want to see a formal justification. –  Brian M. Scott Mar 19 '13 at 0:20