Cartesian product of separable metric spaces

[Andeweld]

Dear readers,

Let $X$ be the product space of a countable family $\{X_n:n\in\mathbb{N}\}$ of separable metric spaces.
If $X$ is endowed with the product topology, we know that it is again separable. Are there other topologies for $X$ such that is separable? Is there a natural metric on $X$ such that $X$ is separable and therefore have a countable base?

The general question is under what conditions on the product space $X$ the following conclusion holds:
"For any topological base $\mathcal{B}$ in $X$, the open subsets of $X$ are countable unions of sets in $\mathcal{B}$"

Thnx

 

[Eynstone]

Let d_n be the metric on Xn. Then d = (sum over n) {(dn/dn+1)/2^n} defines such a metric on X.
The property, however, owes more to the countability axioms. If each of a finite family of spaces has a countable basis, so must the product space in product topology.

 

[Greg Bernhardt]

for the interesting question! As a fellow math enthusiast, I will try my best to provide some insight into this topic.

To answer your first question, there are indeed other topologies for X in which it is separable. One example is the box topology, where a basis for the open sets is given by the Cartesian product of open sets in each X_n. This topology is also separable, as we can take the countable union of open sets from the basis to form a countable dense subset in X.

As for your second question, there is indeed a natural metric on X that makes it separable. This is known as the product metric, which is defined as follows: for any x = (x_1, x_2, ...) and y = (y_1, y_2, ...) in X, we define d(x,y) = max{d(x_n, y_n) : n \in \mathbb{N}}. It can be shown that this metric makes X separable, as we can take the countable set of points with rational coordinates in each X_n to form a countable dense subset in X.

To address the general question, the condition for the conclusion to hold is that X is a second countable space. This means that there exists a countable basis for the topology of X. In the case of a product space, this is satisfied if all the X_n's are second countable. This is because the product of countable sets is also countable, and thus we can take the countable union of sets in the basis for each X_n to form a countable basis for X.

I hope this helps to clarify some of your questions. Keep exploring the fascinating world of topology!