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Can the closed interval [0, 1] be expressed as the union of a sequence of disjoint closed subintervals each of length smaller than 1? Explain.

I have been trying to figure out this problem, and I do not know where to even begin.

A.Fried
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1 Answers1

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No.

Supposing $I_n$ were such a sequence of closed intervals, consider the union of their boundary points $B=\bigcup\partial I_n$. Can you show that $B$, as a subspace of $[0, 1]$, is a complete metric space? What can you say about each set $\partial I_n$ as a subset of $B$ (HINT: think about density)? Do you know a theorem about how the countable union of such sets must behave in a complete metric space?

Noah Schweber
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  • Nice. Much better than the grubby proof I'd have given. Let's see.. $B$ is a closed set (therefore complete though I don't see how that helps). $\partial I_n$ is finite, so $B$ is countable (don't know what "as a subset of $B$" means here). Plainly, $B$ has no isolated points. So $B$ is a countable perfect set. The theorem we need is that every nonempty perfect subset of $\mathbb R$ has the cardinality of the continuum. Am I right? – bof Dec 07 '16 at 23:00
  • @bof That's one way to do it, although that doesn't extend to closed sets. I was thinking of something more Baire-bones. – Noah Schweber Dec 07 '16 at 23:02