Can $\mathbb R$ be written as the disjoint union of (uncountably many) closed intervals?

Question

From this post: http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/

We know that $\mathbb R$ can't be written as the countable union of disjoint closed intervals. Can we do it if we allow uncountably many intervals? There doesn't seem to be a nice way to construct such a cover, since if we choose a closed interval, we split the line into two remaining pieces to cover that re still homeomorphic to the original real line. But on the other hand, I feel as if it should be possible, since with the possibility of uncountably many intervals, you should be able to find a way to cover everything.

It is possible if you allow degenerate intervals $[a,a]$, of course. — hmakholm left over Monica, Oct 17 '14 at 22:02

Kaj Hansen · Accepted Answer · 2020-10-16T08:16:38.690

14

No.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, within each interval $[a_i, b_i]$, we can select a single $q_i \in \mathbb{Q}$ with which to "label" the interval. The set of $q_i$'s is a subset of $\mathbb{Q}$, which is countable.

edited Oct 16 '20 at 08:16

answered Oct 17 '14 at 22:01

Kaj Hansen

33,511

Short and sweet. Thanks! – Nishant Oct 17 '14 at 22:04
Glad I could help! – Kaj Hansen Oct 17 '14 at 22:04
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This shows more: it is not possible to have any uncountable family of pairwise disjoint closed subintervals in $\mathbb{R}$. The same holds with closed balls in any separable metric space. :) – Pete L. Clark Oct 19 '14 at 05:22
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@PeteL.Clark, indeed it does! Basically the same problem came up for $\mathbb{R}^n$ in class on Friday. :) – Kaj Hansen Oct 19 '14 at 05:25

Can $\mathbb R$ be written as the disjoint union of (uncountably many) closed intervals?

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