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?
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.