Cardinality of a dense open set In a compact space of cardinal $2^{\aleph_0}$, can a dense open subset be countable?
 A: A dense open subset of a compact space of cardinality $2^{2^\omega}$ can even be countable: an example is $\beta\Bbb N$ (or $\beta\omega$, if you prefer), the Čech-Stone compactification of the natural numbers. This answer to an earlier question shows that $2^{2^\omega}$ is the maximum possible cardinality of a separable Hausdorff space, and this answer shows that $|\beta\Bbb N|=2^{2^\omega}$.
Added: An example of cardinality $2^\omega$ can be built in $\Bbb R^2$ as follows. For $n\in\Bbb Z^+$ let $$D_n=\left\{\left\langle\frac{2k+1}{2^n},\frac1{2^n}\right\rangle:k=0,\dots,2^{n-1}-1\right\}\;,$$ and let $$D=\bigcup_{n\in\Bbb Z^+}D_n\;.$$ Let $X=D\cup[0,1]$ as a subspace of $\Bbb R^2$ with the usual topology. Then $D$ is a countable dense set of isolated points in $X$, which is compact and of cardinality $2^\omega$.
A: Another example: Consider the homeomorphism $\varphi: \mathbb{R}^n \to B_{1}(0)$ of $\mathbb{R}^n$ with the open unit ball $B_1(0)$ given by $\varphi(x) = \frac{x}{1+\|x\|}$ and let $X$ be the closure of $\varphi(\mathbb{Z}^n)$ in the closed unit ball of $\mathbb{R}^n$. Then $X = \varphi(\mathbb{Z}^n) \cup S^{n-1}$ is compact, and assuming $n \geq 2$ it has cardinality $2^{\aleph_0}$. Moreover, $\varphi(\mathbb{Z}^n)$ is countable, open and dense in $X$.
