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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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Do you allow degenerate intervals like $[a,a] = \{ a \}$? –  Arthur Fischer Nov 30 '12 at 9:58
    
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! –  Arthur Fischer Nov 30 '12 at 10:19

2 Answers 2

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:

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

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.

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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
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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

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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