Mathematics Asked on December 25, 2021
In Willard, it’s given that, for Hausdorff non-singleton spaces –
$prod_{alphain A}X_alpha$ is separable iff $X_alpha$ is separable $forallalphain A$ and $|A|lemathfrak{c}$
From reading the proof, I found that we could prove $prod_{alphain A}X_alpha$ is separable $implies$ $X_alpha$ is separable without assuming $X_alpha$ to be Hausdorff. Hausdorff-ness of $prod_{alphain A}X_alpha$ was only used to show $|A|lemathfrak{c}$.
So, is there an example of a non-Hausdorff product space $prod_{alphain A}X_alpha$ such that $prod_{alphain A}X_alpha$ 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 $prod_{alphain A}X_alpha$ which satisfies the above condition?
If not, then a non-Hausdorff $T_0$ product space?
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.
Answered by Henno Brandsma on December 25, 2021
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.
Answered by Cronus on December 25, 2021
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