If $I_n = \{i \in \mathbb{N} : 1 \leq i \leq n\}$ and if $\mathcal{X}_n=\{(X_i,d_i) : i \in I_n\}$ is a finite family of metric spaces, we know that we can make their product $X = \prod_{i \in I_n}X_i$ a metric space by setting $d: X \times X \to \mathbb{R}$ to be simply $d((p_1,\dots,p_n),(q_1,\dots,q_n))=|(d_1(p_1,q_1),\dots,d_n(p_n,q_n))|$ where $|\cdot|$ is some norm in $\mathbb{R}^n$ (I've seem this done mostly with the $p$-norm).

Now, what if we do the following. Let $\Lambda$ be an arbitrary indexing set (can be even uncountable), then we have a family $\mathcal{X}_\Lambda = \{(X_\lambda, d_\lambda) : \lambda \in \Lambda\}$ of metric spaces. We define their product to be:

$$X = \prod_{\lambda \in \Lambda}X_\lambda = \left\{f : \Lambda \to \bigcup_{\lambda \in \Lambda}X_\lambda : f(\lambda) \in X_\lambda, \forall \lambda \in \Lambda\right\},$$

is it possible then to present a natural metric in $X$ in terms of the metrics in each $X_\lambda$? In other words: "is the arbitrary product of metric spaces a metric space again?"

I've seem a similar question here, but I think this isn't duplicate, since that question was regarding aspects about the category of metric spaces.

Thanks very much in advance!

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    $\begingroup$ So, just to be clear, you don't care whether the resulting metric space is the actual product of the $X_\lambda$ in the category of metric spaces, you just want to put a "natural" metric on their product-as-topological-spaces? $\endgroup$ – Zev Chonoles Aug 11 '13 at 2:11
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    $\begingroup$ Yes, it was what I was thinking about. But as I see from @xyzzyx answer this isn't possible. $\endgroup$ – user1620696 Aug 11 '13 at 2:14
  • $\begingroup$ If you allow the distance between points to equal $\infty$, I believe the answer is 'yes'. But, I could be completely wrong. $\endgroup$ – goblin Aug 11 '13 at 5:58
  • $\begingroup$ Every set can be endowed with the discrete metric. So the answer depends on what you mean by natural. xyzxyz's answer shows that the answer is no if natural includes inducing the product topology. $\endgroup$ – wildildildlife Aug 12 '13 at 15:19

You could define a metric space $(X,d)$ by the rule $d(x,y)=\sup_{\lambda\in \Lambda}{d_{\lambda}(x_{\lambda},y_{\lambda})}$.

Exercise 1: Is $(X,d)$ a metric space, i.e. are the axioms of a metric space satisfied? (I'm not saying whether they or or they aren't; it's an exercise to figure it out!)

It's also pertinent which topology you'd like to induce on the product. If $\Lambda$ is uncountable, then there's no guarantee that the usual product topology on $X$ is metrizable. E.g., see xyzzyx's excellent counterexample.

I hope this helps!

Edit: See Qiaochu's excellent answer below in which he discusses the $d$ I defined above in more detail!

  • $\begingroup$ Did you mean sup rather than inf? $\endgroup$ – Moishe Kohan Aug 11 '13 at 3:29
  • $\begingroup$ Thanks @studiosus for the correction! Yes, I meant $\sup$. I've edited my answer. $\endgroup$ – Amitesh Datta Aug 11 '13 at 5:56
  • $\begingroup$ Thanks for your answer @AmiteshDatta, it helped a lot. Also, the counterexample of $\mathbb{R}^{\mathbb{R}}$ also helped a lot understanding this. Thanks again for the help. $\endgroup$ – user1620696 Aug 12 '13 at 1:40

No. Every metric space is first-countable, and e.g. $\mathbb{R}^\mathbb{R}$ is not first-countable.

  • $\begingroup$ Did you mean 1st countable? $\endgroup$ – Moishe Kohan Aug 11 '13 at 2:13
  • $\begingroup$ @studiosus yeah, sorry. $\endgroup$ – xyzzyz Aug 11 '13 at 2:14
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    $\begingroup$ With which topology is it not first-countable? Does this necessarily contradict @user1620696's wishes? $\endgroup$ – Jonathan Y. Aug 11 '13 at 23:03
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    $\begingroup$ When we're talking about products of topological spaces, it is usually assumed that the topology is a product topology, unless explicitly stated otherwise. $\endgroup$ – xyzzyz Aug 12 '13 at 23:23

Going to spoil the surprise: the metric Amitesh Datta defined is the categorical product in a suitable category of metric spaces, but

  • the morphisms in this category need to be weak contractions: that is, if $f : M \to N$ is such a morphism, then we need $d_N(f(x), f(y)) \le d_M(x, y)$;
  • the metrics in this category need to take the value $\infty$; and
  • the induced topology is not the product topology.

The first requirement makes isomorphisms in this category the same as isometric isomorphisms, which is not true of the usual category of metric spaces and continuous maps (which should really be called the category of metrizable spaces). The usual category of metric spaces and continuous maps is somewhat poorly behaved, e.g. categorical constructions in this category do not come with natural metrics even when they exist.

The second requirement is very natural, e.g. it also allows colimits in a nice way. See Lawvere metric space for details and this blog post for some more details about the special case of Banach spaces (although here we don't allow infinite values of the metric and so we need to restrict our attention to a bounded version of the product).


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