The Cantor ternary set is totally disconnected

Question

A set $S$ in a metric space $X$ is called totally disconnected if for any distinct $x,y\in S$, there exists separated sets $A$ and $B$ with $x\in A$, $y\in B$ and $S=A \cup B$.

Let $C=\bigcap_{n=1}^\infty C_n$ be the Cantor ternary 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.

Answer 1

I begin by assuming that by the $C_n$ you mean the usual closed sets whose intersection is the Cantor ternary set:

• $C_1 = [0,1]$;
• $C_2 = [0,\frac 13] \cup [ \frac 23 , 1 ]$;
• $C_3 = [0,\frac 19] \cup [\frac 29,\frac 13] \cup [\frac 23,\frac 79] \cup [\frac 89, 1]$;
• etc.

Note that for each $n$ the set $C_n$ is made up of disjoint closed intervals of length $3^{-(n-1)}$, and that these intervals are therefore separated from each other. Also note that if $x,y \in C_n$ are such that $3^{-(n-1)} < |x-y|$, then $x,y$ belong to different closed intervals making up $C_n$.

Given distinct $x,y \in C$ essentially by the Archimedean property there must be an $n$ such that $3^{-(n-1)} < |x-y|$, and as $x,y \in C_n$ it follows that they belong to different closed intervals making up $C_n$. Let $I$ be the closed interval in $C_n$ containing $x$. It follows that $x \in C \cap I$ and $y \in C \setminus I$, and these sets are separated.

Answer 2

Let $S$ be the set of open intervals that are removed from $[0,1]$ to produce the Cantor set $C.$ The members of S are pair-wise disjoint, and the sum of their lengths is $\frac {1}{3}\sum_{n=0}^{\infty}(\frac {2}{3})^n=1.$

So no non-empty open interval $J\subset [0,1]$ is a subset of $C.$ Otherwise $T=\{J\}\cup S$ would be a set of pair-wise disjoint open intervals, each a subset of $[0,1],$ with the sum of the lengths of the members of $T$ being greater than $1.$

So when $x,y\in C$ with $x<y$ we have $(x,y)\not \subset C,$ so there exists $z\in (x,y)$ \ $C.$ So let $A=[0,z)\cap C$ and $B=(z,1]\cap C.$