# Countable product of Polish spaces

Let $X$ be a Polish space and let us fix a metric $d$ on it so that $(X,d)$ is a separable complete metric space. Let $Y = X^{\Bbb N}$ be the product space endowed with the product topology. Is it true that $Y$ is a separable space? If not, would it be true in case $X = \Bbb R$?

I also wonder about a metric consistent with the product topology on $Y$. I used to think that $$\rho(y',y'') = \sup\limits_{n\geq 1}d(y'_n,y''_n)$$ is the needed one (here $y' = (y'_1,\dots,y'_n,\dots)\in X^{\Bbb N_0} = Y$) but now I am afraid that it is not the case.

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If you google for countable product separable, you will find several results, such as this blog post or this post from Topology Q+A Board. This is even true for $\mathfrak c$-many spaces Hewitt-Marczewski-Pondiczery theorem,see this question for some references. – Martin Sleziak Dec 2 '12 at 10:55

I will just comment on the metric you have come up with.

First of all, note that the formula you have provided may take value $\infty$ if the metrics under consideration are not (uniformly) bounded. (As an example, let $\vec{x} = ( n )_n$ and $\vec{y} = ( -n )_n$ in where $X_n = \mathbb{R}$ for all $n$. Then $d ( \vec{x} , \vec{y} ) = \sup_n | n - (-n) | = \sup_n 2n = \infty$.)

Secondly, note that the balls under this metric will almost certainly not be open in the usual product topology. Again taking $X_n = \mathbb{R}$ for all $n$, and letting $\vec{0}$ denote the origin, the $1$-ball about $\vec{0}$ consists of sequences $\vec{x}$ such that $| x_n | < 1$ for all $n$ (but not all such sequences, as $\vec{x} = ( 1 - \frac{1}{n} )_n$ is distance $1$ from $\vec{0}$). However, if $U = \prod_n U_n$ is a basic open set in the product topology then there is an $N$ such that $U_n = \mathbb{R}$ for all $n \geq N$. Thus the $1$-ball about $\vec{0}$ includes no basic open sets!

Any errors in previous versions of this answer are solely my fault!

Your formula does define a metric, let's call it $\rho$, on $\prod_n X_n$, though it certainly does not generate the product topology. If $X_n = [-1,1]$ for all $n$ with the usual metric, then the $\rho$-metric topology appears to coincide with the topology $[-1,1]^\mathbb{N}$ inherits as a subspace of the $\ell^\infty$-space. (A big thank-you to Nate Eldredge for pointing this out!)

In general, the $\rho$-metric topology is finer than the product topology (and they only coincide if all but finitely many factors are trivial). Some more or less simple observations are the following:

• If each $X_n$ is discrete, then the $\rho$-metric topology is discrete.

• A consequence of the above is that the $\rho$-metric topology may fail to be compact even if all factors are compact.

Carlos R. Borges, The sup metric on infinite products, Bull. Austral. Math. Soc. v.44 (1991), pp.461-466, MR1138022, link

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I worried myself for a moment there, but I was right: when $\rho$ is a metric, it generates the box topology. – Brian M. Scott Dec 2 '12 at 11:12
@Brian: Oh, yes. Box topology. I hadn't even thought of that. Thanks! – arjafi Dec 2 '12 at 11:24
@BrianM.Scott: No, I'm pretty sure it's not the box topology. For instance, if $X = [-1,1]$ with its usual topology, then $(Y, \rho)$ is the unit ball of $\ell^\infty$. The set $\prod_n (-1/n, 1/n)$ is open in the box topology but definitely not in $\rho$. – Nate Eldredge Dec 2 '12 at 13:53
@Nate: Yikes!${}$ – arjafi Dec 2 '12 at 15:02
Arthur, I would blame Nietzsche. :-) – Asaf Karagila Dec 2 '12 at 15:30

Let $X$ be a separable space with countable dense subset $H$. Then $X^{\mathbb N}$ is a separable space. Fix a point $y \in X$ and consider the set $D$ of all elements $z \in X^{\mathbb N}$ such that $z_n=y$ for all but finitely many $n$ and for those finitely many $n$, $z_n \in H$. This is a countable set. Now for any open basis element $U$ of $X^{\mathbb N}$ we can construct an element of $D$ inside it, since $U_n=X$ for all but finitely many $n$.

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Thanks! Do you know, whether the metric I mentioned is the right one? – S.D. Dec 2 '12 at 10:59
You need to be more careful, as shown in Arthur's answer. See here for the correct metric. – JSchlather Dec 2 '12 at 11:14