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.

  • $\begingroup$ Is it obvious that a closed set has either $\le\aleph_0$ or $\mathfrak c$ elements? $\endgroup$ – hmakholm left over Monica Apr 5 '16 at 7:00
  • 1
    $\begingroup$ @HenningMakholm No, it isn't, but it is something that has been discussed on this site several times. $\endgroup$ – Andrés E. Caicedo Apr 5 '16 at 13:17
  • $\begingroup$ @HenningMakholm: It is not obvious but is well-known; it follows from the Cantor-Bendixson theorem and the fact that perfect sets must have cardinality $\mathfrak{c}$. $\endgroup$ – Eric Wofsey Apr 5 '16 at 16:49

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