Existence of Hausdorff space comprised of disjoint dense subsets

Question

Problem $13H$ of Willard states -

For any set $X$, there is a Hausdorff space $Y$ which is the union of a collection $\{Y_x:x\in X\}$ of disjoint subsets dense in $Y$.

I have no idea how to do this. Any help is appreciated!

Answer 1

Here’s a sketch of one way; I’ve left a few details to be filled in or checked.

Let $Z$ be a Hausdorff space that is the union of two disjoint dense subsets, $D_0$ and $D_1$. For each $x\in X$ let $Z_x$ be a copy of $Z$, and let $Y=\prod_{x\in X}Z_x$. For each $A\subseteq X$ let $\chi_A$ be the indicator function of $A$, and let

$$D_A=\prod_{x\in X}D_{\chi_A(x)}\;.$$

Then $\{D_A:A\subseteq X\}$ is a family of $2^{|X|}$ pairwise disjoint dense subsets of $Y$ whose union is $Y$, which is more than we need: just keep the union of $|X|$ of them. For instance, use

$$Y_0=\bigcup_{x\in X}D_{\{x\}}$$

Answer 2

For any infinite cardinal $\kappa$, consider the space $Y=\{0,1\}^\kappa$. Note that every nonempty open subset of $Y$ has cardinality $2^{\kappa}$, and the usual basis for the product topology consists of $\kappa$ open sets (since any basic open set involves only finitely many of the coordinates).

Now, you can get a dense subset of size $\kappa$ of $Y$ by just picking one point from each nonempty basic open set. Doing this repeatedly in a transfinite recursion of length $2^\kappa$, you can get $2^\kappa$ disjoint dense subsets of $Y$. (It will always be possible to choose new points from each nonempty basic open set since at each stage of the recursion, you have picked fewer than $2^\kappa$ points so far.) So, if you pick $\kappa$ such that $2^\kappa\geq|X|$, you can partition $Y$ into $|X|$ dense subsets.