# The Cantor set, II

This is post # 7 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

In the previous post The Cantor set, I, we constructed the “middle third” Cantor set and we showed that it has cardinality continuum. In this post we discuss some basic topological properties of the Cantor set.

The Cantor set we defined in the previous post is the “middle third” Cantor set, the traditional version found in many texts. In this “middle third” recipe, the open interval in the middle third of $[0,1]$ is taken out. Two closed intervals remain. In each of the succeeding stages, the open interval in the middle third of any remaining closed interval is removed. After the inductive process is completed, the remaining points form the Cantor set. Note that the open intervals removed in this process have a total length of $1$ since $\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots =1$. Yet the points remained have the same cardinality as the unit interval $[0,1]$.

There are other striking properties about the Cantor set. For example, the Cantor set contains no open intervals $(a,b)$. This makes the Cantor set a very “thin” set (a nowhere dense set). On the other hand, the Cantor set has no isolated points since every point of the Cantor set is a limit point of itself.

Cantor Set Preliminary
Let $C$ be the Cantor set as defined in the previous post. Recall that $C=\bigcap \limits_{n=0}^{\infty}A_n$ where

$A_0=[0,1]$
$A_1=B_0 \cup B_1$
$A_2=B_{00} \cup B_{01} \cup B_{10} \cup B_{11}$
$A_3=B_{000} \cup B_{001} \cup B_{010} \cup B_{011} \cup B_{100} \cup B_{101} \cup B_{110} \cup B_{111}$
and so on.

Each $B_g$, where $g$ is a string of zeros and ones of finite length, is a closed interval and is obtained by removing the middle third of the previous closed interval. For example, $B_{010}$ and $B_{011}$ are obtained by removing the middle third of the closed interval $B_{01}$. Refer to the previous post The Cantor set, I for more information about the construction.

Each point $y \in C$ is the single point in the intersection of a nested decreasing sequence of $B_g$, i,e, $\left\{y\right\}=\bigcap \limits_{n=1}^{\infty}B_{g_n}$ where $g_i$ and $g_j$ agree on the first $i$ many coordinates if $i. The $g_n$ can be obtained by tracing the construction steps. For example, $y \in B_0$ or $y \in B_1$. Say $y \in B_0$. Then $y \in B_{00}$ or $y \in B_{01}$. Say, $y \in B_{01}$. Now $y \in B_{010}$ or $y \in B_{011}$. Say, $y \in B_{010}$. Thus, we have:

$\displaystyle y \in B_0 \cap B_{01} \cap B_{010} \cap \cdots$

As a result, each $y \in C$ is associated with $h(y)$, which is a sequence of countably infinite many zeros and ones such that the first $n$ coordinates of $h(y)$ agree with $g_n$.

The mapping $h$ in the above paragraph is a bijection between the Cantor set $C$ and $S_{\omega}$, the set of all countably infinite sequences of zeros and ones. If $g \in S_{\omega}$, we let $g_n$ be the string of zeros and ones that agrees with the first $n$ coordinates of $g$. Thus in this notation, for $y \in C$, we have $\left\{y\right\}=\bigcap \limits_{n=1}^{\infty}B_{h(y)_n}$.

To facilitate the subsequent discussion, we have the following lemma.

Lemma

1. For each $y \in C$, the intervals $B_{h(y)_n}$ form a local base at $y$, i.e. for each open interval $(a,b)$ with $y \in (a,b)$, there is some integer $n$ such that $B_{h(y)_n} \subset (a,b)$.
2. Let $D$ be the set of endpoints of the closed intervals $B_{g_n}$ obtained in the construction process. We have $D \subset C$.
3. The set $D$ is dense in $C$.

Proof
Condition 1 follows from the fact that the lengths of the closed intervals $B_{h(y)_n}$ go to zero.

We now prove condition 2. Let $B_{g_n}$ be a closed interval obtained in the construction process and let $s$ be the left endpoint and $t$ be the right endpoint. Both $s$ and $t$ belong to $C$ since:

$\displaystyle \left\{s\right\}=B_{g_n} \cap B_{g_n0} \cap B_{g_n00} \cap B_{g_n000} \cap \cdots$

$\displaystyle \left\{t\right\}=B_{g_n} \cap B_{g_n1} \cap B_{g_n11} \cap B_{g_n111} \cap \cdots$

The fact that $D$ is dense in $C$ follows from condition 1. Every open interval containing a point of $C$ has to contain one closed interval $B_{g_n}$ and thus the endpoints of $B_{g_n}$. $\blacksquare$

Main Discussion
We discuss the following proeprties:

1. The Cantor set $C$ is a compact subset of $[0,1]$.
2. The Cantor set $C$ is a perfect set.
3. $[0,1]-C$, the complement of $C$, is dense in $[0,1]$.
4. The Cantor set $C$ contains no open interval.
5. The Cantor set $C$ is nowhere dense in $[0,1]$.
6. The Cantor set $C$ is totally disconnected.

Discussion of 1
In the real line or in any other topological space, the intersection of closded sets is always a closed set. The Cantor set $C$ is the intersection of $A_n$ where each $A_n$ is a closed set (it being a finite union of closed intervals). Thus the Cantor set $C$ is a closed subset of $[0,1]$. Being a closed and bounded subset of the real line, $C$ is compact. $\blacksquare$

Discussion of 2
Recall that a subset $W$ of the real line is a perfect set if it is closed and every point of $W$ is a limit point of $W$. Let $y \in C$. Let $(a,b)$ be an open interval of the real line such that $y \in (a,b)$. We want to show that there are infinitely many points of $C$ inside $(a,b)$. This following from condition 1 of the lemma. The interval $(a,b)$ contains $B_{g_n}$ which would contain infinitely many points in the Cantor set. Note that the construction process of the Cantor set would continue within the interval $B_{g_n}$. $\blacksquare$

Discussion of 3
This also follows from condition 1 of the lemma. We need to demonstrate that every point in $[0,1]$ is either a point of $[0,1]-C$ or a limit point of $[0,1]-C$. Let $y \in [0,1]$. If $y \in [0,1]-C$, then we are done. So suppose $y \in C$. By condition 1 of the lemma, every open interval containing $y$ contains some closed interval $B_{g_n}$. Since the middle third of $B_{g_n}$ is removed in the construction process, we see that $(a,b) \cap ([0,1]-C) \ne \phi$. $\blacksquare$

Discussion of 4
This is a corollary of 3. Suppose we have $(a,b) \subset C$. By 3, $(a,b)$ contains points of the removed open intervals, which is a contraction. $\blacksquare$

Discussion of 5
Let $A \subset \mathbb{R}$. The set $A$ is nowhere dense in $\mathbb{R}$ if $A$ is dense in no open interval of $\mathbb{R}$, or equivalently, for each open interval $(a,b)$, there is some subinterval $(c,d) \subset (a,b)$ such that $(c,d) \cap A=\phi$. To say that $A$ is nowhere dense in $[0,1]$, we only need to concern with the open intervals that have points in $[0,1]$.

This is a corollary of condition 4. For each open interval $(a,b)$ in $[0,1]$, it cannot be a subset of $C$. So there must some point $x \in (a,b)$ such that $x \notin C$. But the Cantor set $C$ is a closed subset of $[0,1]$. Then there must be some $(c,d) \subset (a,b)$ such that $x \in (c,d)$ and $(c,d) \cap C=\phi$. $\blacksquare$

Discussion of 6
The Cantor set is totally disconnected. A subset $A$ of the real line $\mathbb{R}$ is a disconnected set if there are disjoint open sets $U$ and $V$, both containing points of $A$, such that $A=(U \cap A) \cup (V \cap A)$. A set $A$ is connected if it is not disconnected.

The Cantor set $C$ is disconnected. Note that $C=([0,0.5) \cap C) \cup (0.5,1] \cap C)$.

A set $A \subset \mathbb{R}$ is totally disconnected if the only connected subsets of $A$ are the singleton sets. We show that any subset of the Cantor set with at least two points is not connected. Let $A \subset C$ with $x,y \in A$ where $x. Since $C$ is nowhere dense, there exists $(c,d) \subset (x,y)$ such that $(c,d)$ has no points of $C$. Let $t=0.5(c+d)$. Then we have:

$\displaystyle A=([0,t) \cap A) \cup ((t,1] \cap A)$