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 ?
Is $[0,1]$ a union of family of disjoint closed intervals?
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3But $[a,a]$ _is_ an [interval](http://mathworld.wolfram.com/Interval.html)! – 2012-11-30
2 Answers
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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6@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. – 2012-11-30
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.