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.


3 Answers 3


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.

  • 3
    $\begingroup$ 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...) $\endgroup$ 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 $x<y$ we have $(x,y)\not \subset C,$ so there exists $z\in (x,y)$ \ $C.$ So let $A=[0,z)\cap C$ and $B=(z,1]\cap C.$


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, $card(A) \le 1$. $\blacksquare$


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