# The Cantor ternary set is totally disconnected

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.

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.

• Good answer, thanks! How does one show that $A=C\cap{I}$ and $B=C\setminus{I}$ are disjoint from the other's closure (I know how to show that $\bar{A}$ is disjoint from $B$, but not the other way around...) Jun 25, 2018 at 0:44

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 $$\frac {1}{3}\sum_{n=0}^{\infty}\left(\frac {2}{3}\right)^n=1.$$

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 $$x we have $$(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.$$

If $$\underline D$$ is Cantor Discontinuum, then we have that $$D = \bigcap D_m, m \in N$$, where: $$D_m=\{ \sum_{n=1}^{\infty} \frac{x_n}{3^n} \ | \ (x_n) -\text{ sequence in } \{0,1,2\} \text{ with } \{x_1,x_2,...,x_m\} \subset \{0,2\} \}$$

Since $$\underline D$$ is equipped with a subspace topology of $$\underline ℝ$$, then any connected set $$A$$ in $$\underline D$$ must be an interval.

If $$card(A) \gt 1 \Rightarrow \exists a, b \in A, \text{(wlog) } a < b$$. So, it must then hold that $$[a,b] \subset A \subset D$$.

Since $$[a,b] \subset D = \bigcap D_m \Rightarrow$$ it must hold that $$\forall m \in N: [a,b] \subset D_m$$.

However, one can always find such $$k \in N : |b-a| \gt \frac{1}{3^k}$$ meaning $$[a,b] \not \subset D_k$$.

Latter implying a contradiction: $$A \not \subset D$$. Therefore, $$card(A) \le 1$$. $$\blacksquare$$