The cantor set $C$ is obtained by repeatedly removing the middle $1/3$, starting from the interval $[0,1]$. Since the number of intervals removed in each step of construction is finite, $[0,1] \backslash C$ is the union of only countable many disjoint intervals.
In a topology book, there is a proof of the fact that a perfect totally disconnected subset on $\Bbb{R}$ is homeomorphic to $C$. In the proof, it is claimed that if a set $A$ is perfect and totally disconnected, then $R\backslash A$ is consisted of countably many disjoint intervals. This is true if $A=C$, but I don't know why it is true for such $A$ in general. Is there a simple explanation for this claim? Can't there be uncountably many intervals in $R\backslash A$?
Best Answer
So first of all every open subset of $\mathbb{R}$ is a countable union of disjoint intervals. I leave this as an exercise.
If you ask why $\mathbb{R}\backslash A$ is not a finite union of intervals then the argument goes as follows: assume that $U=\mathbb{R}\backslash A$ is a finite union of open intervals. Then $A$ is a finite union of closed intervals. Here I treat singletons $\{x\}$ as closed intervals as well. So write down $A=\bigcup_{i=1}^n C_i$ where each $C_i$ is a closed interval and they are pairwise disjoint.
If any of $C_j$ is a singleton then $A$ cannot be perfect. Indeed, let $C_j=\{x\}$. Since there are only finitely many $C_i$'s then some small neighbourhood $V$ of $x$ cannot intersect any of $C_i$ except for $C_j$. Otherewise either $x$ belongs to some bigger interval (contradiction with $C_j$ being a singleton) or there are infinitely many $C_i$ "converging" to $x$ (contradiction with finite number of $C_i$). Therefore $x$ is isolated and so $A$ is not perfect. Contradiction.
It follows that all $C_i$ are proper intervals. But each $C_i$ is a connected component of $A$. That contradicts $A$ being totally disconnected.