"big" Hausdorff space with dense subspace of given cardinality In a topology course we proved the following proposition: > Let $A$ be an infinite set. Then there exists a Hausdorff space $X$ of cardinality $|\mathfrak{P}(\mathfrak{P}(A))|$ which contains a dense subspace of cardinality $|A|$. This proposition has nice consequences - it shows e.g., that there are $|\mathfrak{P}(\mathfrak{P}(A))|$-many (ultra-)filters on any infinite set $A$. However, the only proof I know is horrifically technical. It takes the product topology on $X := A^{\mathfrak{P}(A)}$ (where $A$ is considered to be discrete) and constructs a complicated subspace for which it is not trivial to see, that it is of cardinality $|A|$. I am looking for an elegant proof which is more easy to understand. It is not important for me, whether it uses the same construction or another one, but another construction would of course be particularly interesting :) Thank you in advance!

The **Added** part of this answer contains a proof that there are $2^{2^\kappa}$ ultrafilters on $\kappa$ and hence that $|\beta D|=2^{2^\kappa}$ if $D$ is the discrete space of cardinality $\kappa$. (It wouldn’t surprise me if the construction that you mention in the question is actually a disguised form of the same idea, or at least something similar.)