3

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.

Martin Sleziak
  • 53,687
S.D.
  • 1,417

2 Answers2

6

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.

Some more basic information about this topology can be found in

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

user642796
  • 52,188
4

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 $n$.

JSchlather
  • 15,427