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I'm struggling a little with the idea of compactness. In particular the following question:

If we let $\mathbb{N}_n = \{0,1,2,\dots, n\}$ and equip $\mathbb{N}_n$ with the discrete topology then $\mathbb{N}_n$ is compact if $n$ is finite. Tychonoff's theorem tells us that a product (even an infinite one) of compact spaces is also compact.

So my question is: Is the infinite product space $\Pi_{i = 1}^{\infty}\mathbb{N}_n$ compact where each $\mathbb{N}_n$ is described above?

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    $\begingroup$ Use Tychonoff theorem directly. $\endgroup$ – Hanul Jeon Oct 19 '13 at 15:07
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    $\begingroup$ The short answer is Yes. The long answer is that this is an aspect of the Mystery of Infinite Products. $\endgroup$ – Lubin Oct 19 '13 at 15:44
  • $\begingroup$ You can in fact prove this result without using the Tychonoff theorem at all. $\endgroup$ – Asaf Karagila Oct 19 '13 at 16:01
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As Asaf said in the comments, you don’t need the Tikhonov product theorem to show that this particular product space is compact. The argument is a nice exercise; I’ve done all of the setup below, leaving the payoff part of the argument for you to try if you wish.

Fix an integer $n\ge 2$. Let $I_\varnothing=[0,1]$. If $\sigma$ is a finite sequence of elements of $\Bbb N_n$, and $I_\sigma=[a,b]$ is a closed interval in $[0,1]$, let $\epsilon=\frac{b-a}{2n+1}$. For $k\in\Bbb N_n$ let $x_k=a+k\epsilon$ and $I_{\sigma^\frown k}=[x_{2k},x_{2k+1}]$, where $\sigma^\frown k$ is the finite sequence obtained by appending $k$ to $\sigma$. For $k\in\Bbb N$ let $$C_k=\bigcup\{I_\sigma:|\sigma|=k\}\;,$$ where $|\sigma|$ is the length of the sequence $\sigma$. Each set $C_k$ is the union of finitely many closed intervals in $[0,1]$, so it’s a compact subset of $\Bbb R$. Let $C=\bigcap_{k\in\Bbb N}C_k$; $C_{k+1}\subseteq C_k$ for each $k\in\Bbb N$, so $C$ is a non-empty compact subset of $[0,1]$.

Let $X=\prod_{k\ge 0}\Bbb N_n$; each $x\in X$ is an infinite sequence of elements of $\Bbb N$. For each $k\in\Bbb N$ let $\sigma_k(x)$ be the subsequence of $x$ consisting of the first $k$ terms, and let $$I_x=\bigcap_{k\in\Bbb N}I_{\sigma_k(x)}\;;$$ $I_{\sigma_{k+1}(x)}\subseteq I_{\sigma_k(x)}$ for each $k\in\Bbb N$, so $I_x\ne\varnothing$. $I_{\sigma_k(x)}$ is a closed interval of length $\left(\frac1{2n+1}\right)^k$ for each $k\in\Bbb N$, so $I_x$ must contain just one point (why?); call that point $\varphi(x)$. Now show that the map $\varphi:X\to C$ is a homeomorphism and conclude that $X$ is compact.

$X$ and $C$ are Cantor sets: they are actually homeomorphic to the more familiar middle-thirds Cantor set.

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Each $\mathbb{N}_n$ is compact since the topologies are finite. So the infinite product is compact, by Tychonoff's theorem.

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