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A set $S$ in a metric space $X$ is called totally disconnected if for any $x,y\in S$, there exists separated sets $A$ and $B$ with $x\in A$, $y\in B$ and $S=A \cup B$.Let $C=\cap_{n=1}^\infty C_n$ be the Cantor set.

Given $x,y \in C$ with $x\lt y$, set $\epsilon=y-x$. For each $n\in$ N, $C_n$ consists of a finite union of closed intervals. Explain why there must exist an N large enough so that it is impossible for $x$ and $y$ both to belong to the same closed interval in $C_N$.

I know that the Cantor set is constructed by removing the middle open thirds for each n. And each $C_n$ has $2^n$ closed intervals. As you go on, the closed sets get significantly small, so it's safe to assume that for some N, $x$ and $y$ will be "separated" into two different closed intervals. I'm not sure how to show this formally though.

Any help is appreciated, thanks.

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1 Answer 1

up vote 4 down vote accepted

Hint: For each $n$ you can calculate the length $\ell_n$ of the closed intervals making up $C_n$ (for example, $\ell_1 = 1$, $\ell_2 = \frac{1}{3}$, etc.). Furthermore, if $| x - y | > \ell_n$, then $x$ and $y$ cannot belong to the same interval making up $C_n$.

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So basically for some N, the length of a closed interval $\frac{1}{3^N}$ will be greater than y-x? –  Alti Dec 8 '12 at 3:56
Go it, thank you. –  Alti Dec 8 '12 at 4:07

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