# Example of Separable Product Space with cardinality greater than continuum?

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?

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.

• Ah, of course! I'd completely forgotten about the trivial topology. Also, is there an example of a $T_1$ space satisfying my above condition? If not, then a $T_0$ space? Jul 24, 2020 at 7:53
• @IshanDeo That's a good question. I would look at the product of $\kappa$-many infinite spaces with the cofinite topology and check if it's separable. Jul 24, 2020 at 8:00
• @bof right, right. It was clear to me that this should be an example, but I missed this obvious dense subset... Jul 24, 2020 at 8:18
• But, if we give the product of $κ$-many spaces the co-finite topology, can we then conclude that each of the individual spaces are separable? As that's the main thing I'm trying to find an example about. Jul 24, 2020 at 8:39
• @bof Yes this is true. But is it relevant to the answer? Jul 24, 2020 at 12:05

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.