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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 ?

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PLuS
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    Do you allow degenerate intervals like $[a,a] = { a }$? – user642796 Nov 30 '12 at 09:58
  • no, of course not – PLuS Nov 30 '12 at 10:03
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    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
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    But $[a,a]$ is an interval! – user642796 Nov 30 '12 at 10:19
  • I'm almost sure it doesn't really matter if the intervals are either countable or uncountable. For instance suppose two such intervals in the union are $[a,b]$ and $[c,d]$ such that $b<c$. There is a rational such that $b<x_{n_1}<c$. Inductively, one can always find rationals between $c$ and $d$ with associated disjoint intervals. In the limit $n_j\rightarrow \infty$, the interval becomes of "length" zero, a contradiction because you requiered that the intervals are of positive length and $<1$. – user2820579 Dec 10 '20 at 22:05

2 Answers2

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

$$[0,1]=\bigcup_{x\in[0,1]}[x,x]\;.$$

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.

Brian M. Scott
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  • no, suppose that $[x,x]$ is not allowed! – PLuS Nov 30 '12 at 10:05
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    @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
  • this part: any family of pairwise disjoint closed intervals in $[0,1]$ can include at most countably many non-degenerate intervals. – PLuS Nov 30 '12 at 10:19
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    @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
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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.

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