Example of Separable Product Space with cardinality greater than continuum?

Question

In Willard, it's given that, for Hausdorff non-singleton spaces -

$\prod_{\alpha\in A}X_\alpha$ is separable iff $X_\alpha$ is separable $\forall\alpha\in A$ and $|A|\le\mathfrak{c}$

From reading the proof, I found that we could prove $\prod_{\alpha\in A}X_\alpha$ is separable $\implies$ $X_\alpha$ is separable without assuming $X_\alpha$ to be Hausdorff. Hausdorff-ness of $\prod_{\alpha\in A}X_\alpha$ was only used to show $|A|\le\mathfrak{c}$.

So, is there an example of a non-Hausdorff product space $\prod_{\alpha\in A}X_\alpha$ such that $\prod_{\alpha\in A}X_\alpha$ is separable $\implies$ $X_\alpha$ is separable, $X_\alpha$ is not a singleton, and $|A|>\mathfrak{c}$

EDIT:

Also, is there a non-Hausdorff $T_1$ product space $\prod_{\alpha\in A}X_\alpha$ which satisfies the above condition?

If not, then a non-Hausdorff $T_0$ product space?

Answer 1

If you don't assume Hausdorff-ness, you can pretty much do whatever you want. You can take $X_\alpha$ to be all spaces with the trivial topology, and let $A$ be of as great a cardinality as you want - the product will have the trivial topology, and in particular will be separable.

By the way, for the theorem, you have to assume $X_\alpha$ are moreover not singletons (or rather the theorem says that only $\mathfrak{c}$ of them can have more than one point).

EDIT. Here's an example of a product of $T_1$ spaces. For any cardinality $\kappa$, the product of $\kappa$-many infinite countable spaces with the cofinite topology is separable. As bof pointed out in the comments, the set of constant functions (which is countable) is dense.

Answer 2

If $X$ is Hausdorff, and separable, then $|X| \le 2^\mathfrak{c}$; this is classical (If $D$ is countable and dense, we can show that mapping $x$ to $f(x)=\{A \in \mathscr{P}(D): x \in \overline{A}\} \in \mathscr{P}(\mathscr{P}(D))$ defines an injection from $X$ into a set of size $2^{\mathfrak{c}}$, when $X$ is Hausdorff).

An example where equality is reached is $\{0,1\}^{\Bbb R}$ in the product topology (which is compact Hausdorff, even). For metrisable spaces the bound on $|X|$ is $\mathfrak{c}$, as can be easily seen (that many sequences exist in $D$, a countable dense set).

For $T_1$ spaces there is no such bound as any set $X$ in the cofinite topology is (compact, $T_1$ and) separable, e.g.

Any product of Sierpinski 2-point spaces has a singleton as a dense subset but can be arbitrarily large as well. (This is a $T_0$ but non-$T_1$ example). Any product of cofinite spaces is also separable. So Hausdorff-ness is pretty essential in bounding the size of separable spaces, in products or elsewhere.