TITLE: Continuum hypothesis : Number of connected components of $A^c$ in $\Bbb R$
QUESTION [5 upvotes]: Let $A \subset \Bbb R$ with a countably infinite number of connected component (for the usual topology).
What can be the cardinal of the set of the connected components of $A^c$?
With continuum hypothesis, the situation is clear, there is only two possibilities : 

It's countable ( for exemple take $A = \bigcup_{n\in \Bbb Z} [2n,2n+1]$)
It has the same cardinality as $\Bbb R$ (take $A= \Bbb Q$)

But what if we don't assume continuuum hypothesis? can we obtain situations where the set of the connected components of $A^c$ has intermediate cadinality?

REPLY [5 votes]: Let $P$ be the union of the set of connected components of $A^c$ and the set of connected components of $A$.  Then $P$ is a partition of $\mathbb{R}$ into (possibly degenerate) intervals.  Let $U$ be the union of all the interiors of the elements of $P$ and let $C=U^c$.  Note that $|C|+\aleph_0=|P|+\aleph_0$, since $C$ contains one point for each element of $P$ that is just a point and between $0$ and $2$ points for each element of $P$ that is a nondegenerate interval (depending on how many of the endpoints are in the interval), and there can be only countably many such intervals.  But $C$ is closed in $\mathbb{R}$, so $|C|+\aleph_0$ is either $\aleph_0$ or $\mathfrak{c}$.  Thus $|P|+\aleph_0$ is either $\aleph_0$ or $\mathfrak{c}$.  Since $A$ has countably many connected components, it follows that $A^c$ must have $\mathfrak{c}$ connected components if it has uncountably many connected components.