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I just started my first book on Topology, Introduction to Topology by Bert Mendelson (Second Edition), and the second chapter is about Metric Spaces. The last problem of Section 2, Problem 2.9, is as follows:

Let $(X_n,d_n), n = 1, 2, ...$ be an infinite series of metric spaces. For $x = (x_1, x_2, ...), y = (y_1, y_2, ...) \in \prod_{n=1}^{\infty}X_n$ define $$ d(x,y) = \sum^{\infty}_{n = 1} \frac{d(x_n,y_n)}{2^n}. $$

Prove that $(\prod_{n=1}^{\infty}X_n,d)$ is a metric space.

First off, it seems like a typo that it says $d$ in the sum and not $d_n$.

Here's my beef with the problem: by definition, a metric space has to have a real number as its metric's range, so every possible sum must converge. But the sum doesn't necessarily converge, does it? For example, define for each $X_n$ with $a,b \in X_n$, $d_n(a,a) = 0$ and $d_n(a,b) = 2^n$ for $a \neq b$. If we take $s,t \in \prod_{n=1}^{\infty}X_n$ to be such that $s_n \neq t_n$ for each $n$, then this sum will clearly not converge for $d(s,t)$, right? Is there something I'm missing?

James Hurley
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2 Answers2

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The claim is false, even in the case $$d(x,y)=\sum_{n=1}^\infty\frac{d_n(x_n,y_n)}{2^n}.$$ Take, in this case, $X_n=\mathbb{R}$ and $d_n(x,y)=2^n|x-y|$. Then, each $d_n$ is a metric but $$\begin{split}d((0,0,\ldots),(1,1,\ldots)) &=\sum_{n=1}^\infty\frac{2^n|0-1|}{2^n}\\&=\sum_{n=1}^\infty1\\&=\infty.\end{split}$$ There should be additional assumptions on the $d_n$'s.

RMWGNE96
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The usual assumption is that all metrics $d_n$ are bounded, say by $1$. This ensures the convergence of the series defining the infinite product metric.

This is harmless as $(X,d_n)$ and $(X, \hat{d}_n)$ are always equivalent when $d_n$ is a metric and $\hat{d}_n = \min(d_n, 1)$, the truncated metric.

In this thread I then show that this metric on the countable product induces the product topology.

Henno Brandsma
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