Is the product of hyperconnected separable topological spaces always separable?

Question

The classical Hewitt-Marczewski-Pondiczery theorem says in particular that the product of more than $|\mathbb{R}|$ topological spaces isn't separable, BUT it does the hypotesis that every space has two nonempty disjoint open sets, i.e. all spaces are not hyperconnected (and in the proof this is used very much). So it is natural to asks if and in wich cases the product of hyperconnected separable topological spaces is always separable, or it depends on the cardinality, maybe in a different way from the non hyperconnected case. In particular I'm concerned with understanding in $X^Y$ is separable in the following case: $X$ is an infinite set with the cofinite topology and $Y$ is whatever set.

Answer 1

If $X$ is an infinite set with the cofinite topology, then $X^Y$ is separable for any $Y$. Indeed, let $x_0,x_1,\dots$ be infinitely many distinct points of $X$ and let $c_n\in X^Y$ be the constant function with value $x_n$. Then $\{c_n:n\in\mathbb{N}\}$ is dense in $X^Y$ (any nonempty basic open set can only exclude finitely many of the $x_n$'s on any coordinate).

As for what happens in general, suppose $(X_i)_{i\in I}$ is a family of hyperconnected separable spaces with countable dense subsets $D_i\subseteq X_i$ (which for covenience we assume are infinite). Let us replace $X_i$ with $D_i$ and consider whether $\prod D_i$ is separable (since a countable dense subset of it will also be dense in $\prod X_i$). Identifying each $D_i$ with $\mathbb{N}$, we are reduced to considering products of the form $\mathbb{N}^I$ where on each coordinate $\mathbb{N}$ has some hyperconnected topology (not necessarily the same on each coordinate).

Now suppose there exists a (proper) filter $F$ on $\mathbb{N}$ which contains every nonempty open set on each coordinate. Let $c_n\in\mathbb{N}^I$ be the constant function with value $n$. I claim then that $\{c_n:n\in\mathbb{N}\}$ is dense in $\mathbb{N}^I$. Indeed, given a basic open set $\prod U_i\subseteq\mathbb{N}^I$, the intersection $\bigcap U_i$ is in $F$, since it is a finite intersection of elements of $F$ (since $U_i=\mathbb{N}$ for all but finitely many $i$). In particular, $\bigcap U_i$ is nonempty, and then we see that for any $n\in\bigcap U_i$, $c_n\in \prod U_i$.

More generally, suppose there is a collection of $2^{\aleph_0}$ filters $F_j$ such that for each coordinate $i$, there is some $j$ such that every nonempty open set of the topology on the $i$th coordinate is in $F_j$. Grouping the factors according to this value of $j$, we can write $\mathbb{N}^I$ as a product of spaces $P_j$ where each $P_j$ is separable by the previous paragraph. So $\mathbb{N}^I$ is a product of at most $2^{\aleph_0}$ separable spaces and thus is separable.

Conversely, suppose no such family of $2^{\aleph_0}$ filters exists (for instance, $I$ might be the set of all ultrafilters on $\mathbb{N}$ and each coordinate has its nonempty open sets given by the corresponding ultrafilter). Then I claim $\mathbb{N}^I$ is not separable. Let $F_i$ be the filter on $\mathbb{N}$ generated by the nonempty open sets on the $i$th coordinate (this is a proper filter since each topology is hyperconnected). By hypothesis, we cannot partition $I$ into $2^{\aleph_0}$ pieces such that the union of the $F_i$ on each piece generates a proper filter.

Now suppose we have a countable dense subset $\{s_n:n\in\mathbb{N}\}\subset\mathbb{N}^I$. Define a function $f:I\to \mathbb{N}^\mathbb{N}$ by $f(i)(n)=s_n(i)$. The fibers of $f$ form a partition of $I$ into $2^{\aleph_0}$ pieces, so for some $x\in\mathbb{N}^\mathbb{N}$, the filters $F_i$ for $i\in f^{-1}(\{x\})$ are not compatible. This means we can choose finitely many $i_1,\dots,i_n\in f^{-1}(\{x\})$ and $U_1\in F_{i_1},\dots,U_n\in F_{i_n}$ such that $U_1\cap\dots\cap U_n=\emptyset$. We may moreover assume that each $U_k$ is open in the topology of the $i_k$th coordinate, since the filter $F_{i_k}$ is generated by the nonempty open sets. Now let $U$ be the set of elements of $X^Y$ whose $i_k$th coordinate is in $U_k$ for $k=1,\dots,n$. This is a nonempty open set so it contains $s_m$ for some $m$. But $f(i_k)=x$ for each $k$ so $s_m(i_k)=x(m)$ for each $k$. That is, $x(m)\in U_1\cap\dots\cap U_n$, which is a contradiction.

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So, the upshot is that separability of a product of hyperconnected separable spaces depends not on the cardinality of the index set but on the diversity of the filters on the countable dense subsets determined by the topologies. Note that the nonempty open sets of a hyperconnected topology generate a filter, and extending the topology to include all sets in that filter does not change which subsets are dense (even when you take a product of many such spaces). So, we may as well think of a hyperconnected topology as a filter. Given a family of hyperconnected spaces $X_i$ with countably infinite dense subsets $D_i$ of each one, the topologies on the $D_i$ give a family of filters on $\mathbb{N}$. If those filters can be partitioned into at most $2^{\aleph_0}$ subfamilies where the filters in each subfamily are jointly compatible (i.e., together they generate a proper filter), then the product will be separable. On the other hand, if the filters cannot be partitioned into $2^{\aleph_0}$ subfamilies in this way, then at least the product $\prod D_i$ of the countable dense subsets is not separable. (The product $\prod X_i$ of the whole spaces might still be separable, though, since for instance each $X_i$ might have a dense point that is not in $D_i$.)