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 $|A|>\mathfrak{c}$


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?


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – Ishan Deo
    Jul 24, 2020 at 7:53
  • $\begingroup$ @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. $\endgroup$
    – Cronus
    Jul 24, 2020 at 8:00
  • $\begingroup$ @bof right, right. It was clear to me that this should be an example, but I missed this obvious dense subset... $\endgroup$
    – Cronus
    Jul 24, 2020 at 8:18
  • $\begingroup$ 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. $\endgroup$
    – Ishan Deo
    Jul 24, 2020 at 8:39
  • $\begingroup$ @bof Yes this is true. But is it relevant to the answer? $\endgroup$
    – Cronus
    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.


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