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Let $X$ be a closed and bounded subset of $\mathbb{R}$.

Is it true that $X$ is a finite union of closed intervals in $\mathbb{R}$? (*)

I think that if we choose $X$ is a Cantor set, then $X$ doesn't satisfy (*).

How can I prove this?

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2 Answers

up vote 3 down vote accepted

You are right about Cantor sets in $\Bbb R$: they are not unions of finitely many closed intervals. The easiest way to prove this is to observe that they are all homeomorphic to $\{0,1\}^\omega$, the product of countably infinitely many discrete two-point spaces, and as such are zero-dimensional. In particular, they are totally disconnected: their connected components are singletons. This is not true of any finite union of closed intervals, unless each interval is a singleton, in which case the union is a finite set and therefore obviously not a Cantor set.

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$$X = \left\{ \frac{1}{n} \Bigg|~ n \in \mathbb Z_+ \right\} \cup \{ 0 \} $$

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