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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.)

share|cite|improve this answer
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:… – Cupitor 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! – Cupitor 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

Your Answer


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.