# Show that the countable product of metric spaces is metrizable

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?

• "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." That's not what it's saying. It's saying that the product topology on a countable product of metrisable spaces is metrisable. And it's doing that by giving an example of a metric that does the deed. – kahen Apr 15 '13 at 0:01
• @kahen but these aren't just metrizable spaces, metrics have been chosen for them. – Thoth Apr 15 '13 at 0:14
• The question is very good and this is a classical result. See the Theorem 4.2.2 of the Engelking's book. – Paul Apr 15 '13 at 1:41
• @Thoth: Indeed, but what makes things non-canonical is: 1) the choice of the coefficients $\frac 1 {2^n}$ in front of those fractions (you could have chosen any other convergent series with positive terms); 2) the choice of the function $\frac x {1+x}$ in order to produce bounded distances on each $X_n$ (you could have chosen any other bounded continuous positive function). These two choices are arbitrary, so the whole construction is non-canonical. – Alex M. Jun 23 '18 at 18:17
• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – Luis Felipe Aug 1 at 19:09

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.

• this makes complete sense, thanks for being so detailed. – Thoth Apr 15 '13 at 22:09
• "with an open ball $B_{\rho}(x,r)$, a basic open subset of the ρ-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 $x\in O\subset B_{\rho}(x,r)$." How does it imply that any $\rho$-open ball is open in the product topology? – Babai Dec 8 '14 at 12:49
• @Susobhan Every $\rho$-open ball is itself a $\rho$-open set. So if we have shown what I said, for every $y \in B_\rho(x,r)$ we find some $B_\rho(y, s) \subseteq B_\rho(x,r)$, and then we apply the fact for $B_\rho(y,s)$ and get $O(y)$ product-open inside $B_\rho(y,s)$. But then $B_\rho(x,r) = \cup\{O(y): y \in B_\rho(x,r)\}$ making the whole ball product open. – Henno Brandsma Dec 10 '14 at 12:53
• @HennoBrandsma How did you conclude that $B_{\rho}(x,r)\subset O$? I can see that for all $n\in F$, $y_n\in O_n$, but if $n\not\in F$ then how can you be sure that $y_n\in [0,1]$? – user178318 Mar 26 '17 at 22:41
• @Math_QED If all $X_n$ are complete, under $d_n$, then so is $X$ under $d$. I don't think one needs choice to show that, but I'm no expert on that, Asaf might be better at answering that kind of question... – Henno Brandsma Nov 2 '18 at 15:04