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In this answer, the following fact is assumed:

If $A$ and $B$ are closed subsets of $\mathbb{R}$ such that $A\cup B=\mathbb{R}$, then either $A$ or $B$ contains an interval.

Why is that true? I first thought that we can use measure theory: either $m(A)>0$ and $m(B)>0$ and hence one contain an open interval. But this is false since there are sets of positive measure not containing any interval (e.g. $\mathbb{R}-\mathbb{Q}$). But $\mathbb{R}-\mathbb{Q}$ is not closed, so maybe closed sets with positive measure contain intervals?

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Steph
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  • This need not be true. There are 'fat' cantor sets which are closed sets of positive measure, but do not contain any intervals; see https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set. – SamM Sep 14 '16 at 09:28
  • @SamM how does this show it need not be true? What would the other set be? – user2520938 Sep 14 '16 at 09:29
  • Oh, I apologise, my comment is not very clear. I was answering the question: do there exist closed sets of positive measure which do not contain intervals. (The OP has since made the actual question more clear.) – SamM Sep 14 '16 at 09:30

1 Answers1

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If $A$ is not all of $\mathbb{R}$, then $\mathbb{R}\setminus A$ is a nonempty open set, and so in particular it contains an open interval. But $\mathbb{R}\setminus A\subseteq B$.

(Much less trivially, even if you have countably infinitely many closed sets whose union is $\mathbb{R}$, then by the Baire category theorem one of them must contain an interval.)

Eric Wofsey
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