Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.