Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to this question, $[0,1]$ cannot be written as union of countable disjoint closed sets, is the same true about (uncountable) family of disjoint closed intervals ?

share|cite|improve this question
Do you allow degenerate intervals like $[a,a] = \{ a \}$? – arjafi Nov 30 '12 at 9:58
no, of course not – PLuS Nov 30 '12 at 10:03
You’re free to disallow degenerate intervals, but there’s no of course about it. – Brian M. Scott Nov 30 '12 at 10:04
by of course I mean, that way the answer is trivially no, so I didn't mean it. That's why I dint's ask about closed sets – PLuS Nov 30 '12 at 10:05
But $[a,a]$ is an interval! – arjafi Nov 30 '12 at 10:19
up vote 12 down vote accepted

If you allow degenerate closed intervals, $[0,1]$ can be written as the union of $2^\omega=\mathfrak c$ pairwise disjoint closed intervals:


Since each non-degenerate closed interval contains a non-empty open interval, any family of pairwise disjoint closed intervals in $[0,1]$ can include at most countably many non-degenerate intervals. They cannot cover $[0,1]$, so you’ll need degenerate closed intervals to complete the cover.

share|cite|improve this answer
no, suppose that $[x,x]$ is not allowed! – PLuS Nov 30 '12 at 10:05
@PLuS: I’ve dealt with that: my answer now shows that you can’t decompose $[0,1]$ into more than one pairwise disjoint closed interval unless you use some degenerate intervals. – Brian M. Scott Nov 30 '12 at 10:07
I don't think it is an exact answer, why is your second paragraph correct? – PLuS Nov 30 '12 at 10:16
@PLuS: What part of it don’t you see? – Brian M. Scott Nov 30 '12 at 10:17
@PLuS: Each non-degenerate interval must contain a rational, and there are only countably many rational, so there can be at most countably many pairwise disjoint non-degenerate intervals. More generally, every separable space is ccc, meaning that every family of pairwise disjoint open sets is countable. – Brian M. Scott Nov 30 '12 at 10:20

Suppose $([a_j,b_j])_{j\in J}$ is any family of disjoint, closed, non-degenerate intervals. For every $n\in\Bbb N$ define $$J_n=\{j\in J\mid -n\le a_j,b_j\le n\text{ and }b_j-a_j\ge 1/n\}.$$ Since the intervals are disjoint, every $J_n$ contains less than $2n^2$ elements. Since the intervals are non-degenerate (and bounded), every $j\in J$ is in some $J_n$. Thus $$J=\bigcup_{n\in\Bbb N}J_n$$ is countable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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