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In a compact space of cardinal $2^{\aleph_0}$, can a dense open subset be countable?

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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 $2^\omega$.

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A nitpick: does this really answer the question? It's not quite obvious how to cut down the size from $2^{2^{\omega}}$ to $2^{\omega}$, so a priori it might still be the case that a situation as in the question can't occur. A natural example would be the geometric compactification by adding $S^1$ to an orbit of $SL(2,\mathbb{Z})$ acting on the Poincaré disk. –  t.b. Jun 1 '12 at 8:53
    
@t.b.: It doesn’t literally answer the question; I’m still thinking about that. But since the question may have been a stand-in for ‘How large can such a space be?’, I thought it worth posting. –  Brian M. Scott Jun 1 '12 at 8:57

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 $X$.

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One good nitpick deserves another: you need to specify that $n\ge 2$: if $n=1$, $X$ is just a double-ended sequence. (Basically the same trick that I ended up using.) –  Brian M. Scott Jun 1 '12 at 9:26
    
Okay, you win! Fixed, thanks :) –  t.b. Jun 1 '12 at 9:32
    
We could continue the nitpicking and observe that Hausdorff wasn't specified, so take a cofinite set in a countable space with the cofinite topology... –  t.b. Jun 1 '12 at 10:31
    
Hey, no one said anything about $T_0$; why not start with $\omega+1$ and blow up the limit point to an undifferentiated lump of ... oh, say measurable cardinality? :-) –  Brian M. Scott Jun 1 '12 at 10:42
    
Oh, great! :) At least this example works (if you don't blow the lump up to be too large so as to guarantee $2^{\aleph_0}$ cardinality), contrary to mine... –  t.b. Jun 1 '12 at 10:51

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