Show that the countable product of metric spaces is metrizable

Question

Given a countable collection of metric spaces $\{(X_n,\rho_n)\}_{n=1}^{\infty}$. Form the Cartesian Product of these sets $X=\displaystyle\prod_{n=1}^{\infty}X_n$, and define $\rho:X\times X\rightarrow\mathbb{R}$ by

$$\rho(x,y)=\displaystyle\sum_{n=1}^{\infty}\frac{\rho_n(x_n,y_n)}{2^n[1+\rho_n(x_n,y_n)]}.$$

Show that $\rho$ is a metric on $X$ whose induced topology is equivalent to the product topology on $X$.

So basically what this problem is saying is that there's a canonical way to define a metric on the countable product of metric spaces. I showed in a previous problem that the topology induced by $\rho_n$ is equivalent to that induced by $\frac{\rho_n(x_n,y_n)}{1+\rho_n(x_n,y_n)}$. And thus we can go ahead and just assume that $\rho_n< 1$ for all $n$ and replace our infinite series by

$$\rho(x,y)=\displaystyle\sum_{n=1}^{\infty}\frac{\rho_n(x_n,y_n)}{2^n}.$$

Now comes the interesting part: how should I go about showing that the product topology on $X$ and the topology induced by $\rho$ are equivalent?

The basis for the product topology given to me in my book's definition is that of cartesian products made up the $X_n$ except for finitely many which are $O_n$ for some open subset of $X_n$. However I believe I was able to improve upon this and show that I could decompose these into a basis where the $O_n$ were all open balls induced by their respect $\rho_n$ metric.

For $\rho$ I'm using the basis of open balls that it induces, as I can see no other reasonable choice.

However I can't seem to match these two bases up. There are many different points $\{x_n\}\in X$ which make my infinite series less than a certain value and there is so much freedom in which terms in the series I choose to reduce in size that it seems hopeless to try and fit an open ball induced by $\rho$ into any one of my basis elements for the product topology.

Is there a more appropriate strategy for proving these two topologies are equivalent?

Answer 1

I'll use the $\rho_n$ that are bounded by 1, and $\rho(x,y) = \sum_n \frac{\rho_n(x_n,y_n)}{2^n}$ metric on the product $X = \prod_n X_n$.

Let $O$ be a basic open product set, so $O = \prod_n O_n$, all $O_n$ are open in $X_n$ and where we have a finite set $F \subset \mathbb{N}$ such that $n \notin F$ iff $X_n = O_n$. We want to show it is open in the $\rho$-topology, so pick $x \in O$, and we want to find $r>0$ such that $B_{\rho}(x, r) \subset O$. This would show that all basic product open sets are $\rho$-open, and thus all product open sets are $\rho$-open.

Now, for every $n \in F$, we have that $x_n \in O_n$, which is a (non-trivial) open subset in $X_n$, so we have $r_n > 0$ such that $B_{\rho_n}(x_n, r_n) \subset O_n$, from the fact that the topology on $X_n$ is induced by the metric $\rho_n$. As we have finitely many $r_n$ to consider, we can find $0 < r < 1$ such that $r \le \frac{r_n}{2^n}$ for all $n \in F$.

The claim now is that this $r$ is as required, in the sense that $B_{\rho}(x, r) \subset O$.

To see this, take any $y$ with $\rho(x,y) < r$. For $n \in F$, we know that $\frac{\rho_n(x_n, y_n)}{2^n} \le \rho(x, y) < r \le \frac{r_n}{2^n}$, which implies that for such $n$ we have that $\rho_n(x_n, y_n) < r_n$, and so $y_n \in B_{\rho_n}(x_n, r_n) \subset O_n$. Hence, for all $n \in F$, $y_n \in O_n$, and as the other $O_n$ equal $X_n$ by the form of $O$, we have that indeed $y \in O$, and as $y$ was arbitrary, $B_\rho(x, r) \subset O$, as required.

Now for the other part: we start with an open ball $B_\rho(x,r)$, a basic open subset of the $\rho$-topology, for some arbitrary $x \in X$ and $r>0$, and try to find a basic open subset in the product topology $O$ such that $x \in O \subset B_\rho(x,r)$. This would then show that any $\rho$-open ball is open in the product topology and would show the other inclusion we need: every $\rho$-open set is product open.

The intuition is that the tail of a series like the one that defines $\rho$ is essentially irrelevant (we can get it as small as we like) and this corresponds to the idea that basic open subsets only depend on finitely many non-trivial open sets. So we first pick $N \in \mathbb{N}$ such that $\frac{1}{2^N} < \frac{r}{2}$. This $N$ defines our tail. For $1 \le k \le N$ we consider the open balls $O_k = B_{\rho_k}(x_k, \frac{r}{2N})$, and we set $O_k = X_k$ for $k \ge N+1$.

The claim now is that $O = \prod_k O_k \subset B_\rho(x, r)$, as required. Note that $O$ is indeed a basic open subset in the product topology on $X$ and $x \in O$. To verify the latter claim, we simply estimate: let $y$ be in $O$, then for $k \le N$, $\rho_k(x_k, y_k) < \frac{r}{2N}$, so $$\sum_{k=1}^{N} \frac{\rho_k(x_k,y_k)}{2^k} \le \sum_{k=1}^{N} \rho_k(x_k,y_k) < N\cdot \frac{r}{2N} = \frac{r}{2}\mbox{,}$$ while $$\sum_{k=N+1}^{\infty} \frac{\rho_k(x_k, y_k)}{2^k} \le \sum_{k=N+1}^{\infty} \frac{1}{2^k} = \frac{1}{2^N} < \frac{r}{2}\mbox{.}$$

Putting it together, we indeed get that for $y \in O$ we have $\rho(x,y) < \frac{r}{2} + \frac{r}{2} = r$, as required.