Countable product of metric spaces is metrizable [General Metric]

Question

I know that if we have a countable collection of metric spaces $\{(X_n,\rho_n)\}_{n=1}^{\infty}$ then $X=\Pi^{\infty}_{n=1}X_n$ is a metric space with metric $\rho((x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}})=\sum_{n=1}^{\infty} \dfrac{\rho_n(x_n,y_n)}{2^n [1+\rho_n(x_n,y_n)]}$.

Also I have proved that $X=\Pi^{\infty}_{n=1}X_n$ is a metric space with metric $\rho_\beta((x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}})=\sum_{n=1}^{\infty} \dfrac{\rho_n(x_n,y_n)}{\beta^n [1+\rho_n(x_n,y_n)]}$ where $\beta \in \mathbb{N}$ and $\beta \geq 2$.

But I need to have it from a book or any bibliographic reference, exists some bibliographic reference to this problem with the metric $\rho_\beta$?

You can also take, quite easier, $$\rho((x_n){n \in \mathbb{N}},(y_n){n \in \mathbb{N}})=\sup_{1\le n\le N} \dfrac{\rho_n(x_n,y_n)}{2^n }$$ where $N$ can be finite or $\mathbb N$. — Piquito, Jun 18 '19 at 17:48

score 2 · Answer 1 · answered Jun 18 '19 at 18:45

In the book "Le cours de la APM III, Elements de topologie" by André Revuz and Germaine Revuz (Association des professeurs de Mathématiques de l'Enseignement Public PARIS-1966), you find the statement of exercise 20 (page 38) as follows:

Exercise 20.-Produit dénombrable d'espaces métriques.

1º Si $d$ est une distance definie sur un ensemble $E$, il en est de même de $\delta$ définie par $$\delta(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ et $\delta$ définie la même topologie que $d$.

2º Soit $E_n$, avec $n\in\mathbb N$,une famille dénombrable d'espaces métriques et $E=\prod\{E_n; n\in\mathbb N\}$ leur produit cartésien. A la distance $d_n$ de $E_n$, on associe la distance $\delta_n$ définie en 1º;en on définit sur $E\times E$ (avec $x=\{x_n\}$ et $y=\{y_n\}$ l'application $\delta$ dans $\mathbb R^+$ donnée par $$\delta(x,y)=\sum_{n=1}^{\infty}\frac{1}{2^n}\delta_(x_n,y_n)$$ Montrer que $\delta$ est une distance sur $E$, et que la topologie qu'elle définit est la topologie produit.

This exercise 20 is completely developed on page 172. If you want, I will print the detailed solution in two and a half pages of the book. Where there are also good references is in the analysis books of Jean Dieudonné and those of Laurent Schwartz.

Nice find! As an aside to those who may read this, note that what is written here about $\delta$ ("même" = "same") implies that boundedness of a metric space is $\mathbf{not}$ a topological property (i.e. not preserved under homeomorphism). Indeed, $\delta(v, w) < 1$ for all $v, w \in E$. — Kaj Hansen, Jun 19 '19 at 09:37
@Kaj Hansen: The interesting "additional", which I suppose could interest the OP is that the resulting topology is the product topology (topological and non-metric notion). — Piquito, Jun 19 '19 at 13:44
@KajHansen The OP wanted this not for $2^n$ but $\beta^n$ (though an inspection of the proof will immediately give that), he wanted an explicit reference for that. $2^n$ he already knew... — Henno Brandsma, Jun 19 '19 at 15:28
The OP also does not want to demonstrate the more "theoretical" fact that the resulting topology is the product topology — Piquito, Jun 20 '19 at 13:11

Kaj Hansen · Answer 2 · 2019-06-17T23:06:49.870

Your constructive approach is admirable. If you have no luck in finding your reference, know that there's an indirect way to prove that countable products of metrizable spaces are themselves metrizable:

The Nagata–Smirnov metrization theorem states that a space is metrizable $\iff$ it is Hausdorff, regular and has a countably locally finite basis. It is reasonably straightforward to show that Hausdorffness and regularity are properties preserved under countable products: i.e., the product of a countable collection of spaces sharing one of these properties also has the property. And although I haven't done it myself, I'd wager it's not too horrible to show that this is also the case with the basis condition.

If any of this proves difficult, you could try a similar approach, this time verifying that countable products preserve the necessary and sufficient conditions for metrizability outlined in the Bing metrization theorem: regularity, $T_0$-ness, and the existence of a $\sigma$-discrete basis.

The upshot? It's probably a lot easier to find citations for the metrization theorems.

The OP want specifically this metric, not some metric from a metrisation theorem. That's the whole point. — Henno Brandsma, Jun 17 '19 at 22:15
Yeah, I suspect you're right. I'll wait until the OP logs back in to give him a chance to see this, and then I'll delete if it's definitely of no help. — Kaj Hansen, Jun 17 '19 at 22:52
Yes, I want specifically this metric but all of this things help me, Thank You. — TresTresUno, Jun 18 '19 at 01:39

score -1 · Answer 3 · answered Jun 17 '19 at 06:08

-1

The proof is the same for both cases. All we need is that $\sum_n \frac{1}{\beta^n}$ converges. See my proof here; convergence of such a series is used at one place in the proof.

answered Jun 17 '19 at 06:08

Henno Brandsma

242,131

I know it but I need a bibliographic reference to cite this in a homework. – TresTresUno Jun 17 '19 at 06:11
Maybe google the theorem and cite a reference from google books – Jun 17 '19 at 06:54
@TresTresUno homework does not need bibliographies... Just reprove the my answer with your $\beta^n$ as a lemma. – Henno Brandsma Jun 17 '19 at 22:16
@HennoBrandsma The OP doesn't ask for a proof, in fact they state they already have proved it. – Vsotvep Jun 19 '19 at 20:09
@Vsotvep Then I don't see why one needs a citation for a fact one has just proved. – Henno Brandsma Jun 19 '19 at 21:05
For example to save space in a text that would get too long or would digress too much if the proof of the fact was included. – Vsotvep Jun 19 '19 at 21:06

Countable product of metric spaces is metrizable [General Metric]

3 Answers3