3
$\begingroup$

I want to show that if $C\subseteq \mathbb{R}$ is closed, then $C$ can be written as the union of countably many (or finitely many) disjoint closed intervals.

Note: If $C$ is itself a closed interval then this is trivially true, a bunch of people I have asked say $[0,1]$ is a counterexample, but it is not because $[0,1]=\cup\{[0,1]\}$ which is a finite union of disjoint closed intervals.

I know a similar theorem is true for open intervals.

  • 0
    The corresponding thing that is true is that any closed set is the intersection of countably many closed intervals.2011-10-21
  • 0
    @Joe: What? Any intersection of closed intervals is itself a closed interval.2011-10-21
  • 0
    @Chris Eagle: The Cantor set is an intersection of closed intervals. Just intersect $[0,1]$ with $[0,\frac{1}{3}]\cup [\frac{2}{3},1]$ and so on. But, the Cantor set is not a closed interval.2011-10-21
  • 0
    @Joe: $[0,\frac{1}{3}] \cup [\frac{2}{3}, 1]$ is not a closed interval. What do you actually mean?2011-10-21
  • 0
    @Chris Eagle: My original comment is that if $C$ is a closed set, then there are countably many closed intervals $C_1, \ldots , C_n,\ldots$ such that $C=\cap_i C_i$. So the Cantor set is $[0,1]\cap[0,\frac{1}{3}]\cap[\frac{2}{3},1]\cap\cdots$.2011-10-21
  • 2
    @Joe: That intersection is empty.2011-10-21
  • 0
    @Chris Eagle: No, if you keep removing the middle third of each interval and taking intersections it will still contain the endpoints of the intervals. For instance, 0 and 1 will be in it.2011-10-21
  • 2
    @Joe: As Chris points out, that is not true. The real "dual" is that any closed set can bet written as the countable intersection of sets of the form $(-\infty,a]\cup[b,+\infty)$, which are just the complements of the open intervals.2011-10-21
  • 0
    Sorry, @Joe, but $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$ are disjoint, so Chris is right that your intersection is empty, and not the Cantor set.2011-10-21
  • 0
    @Thomas Andrews: Oh. I consider those closed intervals. Sorry for the confusion. Yes, I see now that my Cantor thing is off. In my head I was thinking $[0,1]\cap ([0,\frac{1}{3}]\cup [\frac{2}{3},1])\cap \ldots$, but wrote out that crap.2011-10-21
  • 0
    Sorry, @Joe, even if you considere $(-\infty,a]$ to be a closed interval, the union if it and $[b,+\infty)$ is no longer a closed interval.2011-10-21
  • 0
    @Thomas Andrews: No doubt about it, I was wrong. My last comment was to explain that I think those particular things are intervals and that I completely botched the Cantor set. No need to say sorry. I often write before I think.2011-10-21
  • 1
    To comment on the problem at hand, it is a common error when first dealing with the idea of closed and opened sets to think that "open sets are to open intervals as closed sets are to closed intervals." This is not the case. It is true that open intervals are open sets, and closed intervals are closed sets, but that is, for most practical purposes, the extent of the similarity.2011-10-21

2 Answers 2

10

The Cantor set is a union of uncountably many points, but contains no closed intervals. So, it cannot be written was the union of countably many disjoint closed intervals.

6

This result is false: any uncountable closed subset of $\mathbb{R}$ with empty interior (e.g. the standard Cantor set) gives a counterexample.