countable product of totally bounded space is totally bounded

Question

Let $ \{ X_i \}_ { i \in \mathbb{N} }$ be a countable collection of metric spaces $(X_i, d_i)$. The product topology on product space $X=\prod_{i=1}^{\infty} X_i$ is equivalent to the metric topology on $X$ induced by $$d(x,y) = \sum_{i}^{\infty} \frac{1}{2^i} \bar{d_i} (x_i,y_i) $$ where $\bar{d_i} (x_i ,y_i) = \min (d_i (x_i, y_i ) , 1) $ is the bounded metric for $d_i $ on $X_i$

I read that the countable product space $(X,d)$ is totally bounded if each factor space $(X_i, d_i)$ is totally bounded, but somewhat hard to prove this fact. Can anybody help me with that proof?

Thanks.

Answer 1

Here’s a ‘nuts-and-bolts’ proof.

Fix $\epsilon>0$; there is an $m\in\Bbb N$ such that $2^{-m}<\frac{\epsilon}4$, and hence $\sum_{k\ge m}2^{-k}<\frac{\epsilon}2$. For $k<m$ Let $F_k$ be a finite subset of $X_k$ such that the open $\frac{\epsilon}2$-balls centred at points of $F_k$ cover $X_k$, and for each $k\ge m$ let $x_k\in X_k$ be arbitrary. Let

$$F=\prod_{k<m}F_k\times\prod_{k\ge m}\{x_k\}\;;$$

clearly $F$ is a finite subset of $X$. Now let $Y=\langle y_k:k\in\Bbb N\rangle$ be arbitrary. For each $k<m$ there is an $x_k\in X_k$ such that $d_k(y_k,x_k)<\frac{\epsilon}2$; let $x=\langle x_k:k\in\Bbb N\rangle\in F$. Then

$$\begin{align*}d(x,y)&=\sum_{k\in\Bbb N}2^{-k}\bar d_k(x_k,y_k)\\ &=\sum_{k<m}2^{-k}\bar d_k(x_k,y_k)+\sum_{k\ge m}2^{-k}\bar d_k(x_k,y_k)\\ &<\sum_{k<m}\frac{\epsilon}{2^{k+1}}+\sum_{k\ge m}2^{-k}\\ &<\frac{\epsilon}2+\frac{\epsilon}2\\ &=\epsilon\;, \end{align*}$$

so

$$X=\bigcup_{x\in F}B_d(x,\epsilon)\;,$$

and $X$ is therefore totally bounded.

Answer 2

In general, an arbitrary (i.e. not necessarily countable) product of compact spaces is compact (this is called Tychonov's theorem, one of the deepest theorems of mathematics). Now, using your notations, if $X_i$ is totally bounded then $\hat {X_i}$ (the metric completion of $X_i$) is compact. By Tikhonov's theorem, then, $\hat X = \widehat{\prod X_i} = \prod \hat X_i$ will also be compact, so the closure of $\prod X_i$ in $\hat X$ will also be compact. But this means exactly that $X$ is totally bounded (compact metric space means complete and totally bounded).

The proof above has the disadvantage that it uses the axiom of choice (which is well hidden inside the proof of Tychonov's theorem). In the case of a countable product (as yours), the full power of Tychonov's theorem is not needed and one can circumvent it, thus also circumventing the use of the axiom of choice. A proof of Tychonov's theorem in this case (which almost proves your own question) can be found here.