0
$\begingroup$

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 $R\backslash A$?

$\endgroup$
  • 1
    $\begingroup$ That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational. $\endgroup$ – freakish Jan 7 at 13:08
  • 1
    $\begingroup$ @freakish Presumably "countable" is used in the sense of "countably infinite". $\endgroup$ – Robert Israel Jan 7 at 13:11
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.