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Consider a fixed set of finite discrete symbols $\mathcal{A}$. Equip $\mathcal{A}$ wit the discrete topology which we denote by $\theta$, and $\mathcal{A}^{\mathbb{Z}^d}$ with the product topology, denoted by $\tau$.

Is then $(\mathcal{A}^{\mathbb{Z}^d},\tau)$ a compact metric space?

I think this has something to do with Tychonoff.

Salamo
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1 Answers1

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It's not a metric space, because you haven't defined a metric. It is, however, a metrizable space: there are metrics on it that generate the product topology. It is homeomorphic to the middle-thirds Cantor set.

One compatible metric is obtained as follows. Let $\varphi:\Bbb N\to\Bbb Z^d$ be any bijection. Let $X=\mathcal{A}^{\Bbb Z^d}$. If $x$ and $y$ are distinct points of $X$, let

$$\delta(x,y)=\min\left\{n\in\Bbb N:x\big(\varphi(n)\big)\ne y\big(\varphi(n)\big)\right\}\;.$$

Then for $x,y\in X$ set

$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 2^{-\delta(x,y)},&\text{if }x\ne y\;; \end{cases}$$

$d$ is a metric on $X$ that generates the topology $\tau$.

Brian M. Scott
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  • One such metric is $d(x,y)=2^{-n}$ with $n=\max\left{m\geqslant 0: \lVert v\rVert_{\infty}<m\implies x_v=y_v\right}$. But why is it compact? Did not understand it yet. – Salamo Dec 24 '14 at 12:03
  • @Salamo It looks like a product of compact spaces, hence compact (by Tychonov's theorem)... – C_M Dec 24 '14 at 12:14
  • @Brian Could you pls expplain HOW your metric generates $\tau$? – Salamo Dec 24 '14 at 12:19
  • @Salamo: As C_M said, it's a product of compact spaces, namely, copies of the finite space $\mathcal{A}$, so by the Tikhonov product theorem it's compact. – Brian M. Scott Dec 24 '14 at 12:19
  • This uses that $\mathcal{A}^{\mathbb{Z}^d}=\prod_{i=1}^d\prod_{j=1}^{\infty}A$? – Salamo Dec 24 '14 at 12:23
  • @Salamo: For each $x\in X$ the sets $$B(x,n)=\left{y\in X:y\big(\varphi^{-1}(k)\big)=x\big(\varphi^{-1}(k)\big)\text{ for all }k\le n\right}$$ is a local base at $x$ in both $\tau$ and the topology generated by $d$. – Brian M. Scott Dec 24 '14 at 12:30
  • @Salamo: There's no need to use a double product. It's simply the product of $\left|\Bbb Z^d\right|$ many copies of $\mathcal{A}$. – Brian M. Scott Dec 24 '14 at 12:33