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\title{The Cantor Set}
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\section{The Cantor Set}
This section describes a set known as the \textbf{Cantor Set} because its construction is due to George Cantor.  The Cantor set exhibits a number of remarkable properties, some of them quite counterintuitive.    
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\subsection{Construction}
To construct the Cantor set, we start with the closed interval $C_0=[0,1]$ and remove the open interval representing the middle third of the set:
\[
C_1=C_0\setminus \left(\frac{1}{3},\frac{2}{3}\right)
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\]
so that
\[
C_1=\left[0,\frac{1}{3}\right] \cup \left[\frac{2}{3},1\right]
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\] 
Now construct $C_2$ by removing the open intervals representing the middle third of each of the two parts of $C_1$:
\[
C_2=C1\setminus \left\{\left(\frac{1}{9},\frac{2}{9}\right) \cup \left(\frac{7}{9},\frac{8}{9}\right)\right\}
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\]
So 
\[
C_2= \left(0,\frac{1}{9}\right) \cup \left(\frac{2}{9},\frac{1}{3}\right) \cup \left(\frac{2}{3},\frac{7}{9}\right) \cup \left(\frac{8}{9},1\right)
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\]
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Now continue the process inductively, that is, construct $C_3$ from $C_2$, $C_4$ from $C_3$, and so on, each time removing the open middle third of each interval in $C_n$ to product $C_{n+1}$.  
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The result is a sequence of sets $C_0,C_1,C_2,\ldots$ in which $C_n$ is the union of $2^n$ closed intervals, each with length $1/3^n$.  Now define the Cantor set to be the intersection of this collection:
\[
C=\bigcap_{n=0}^{\infty} C_n
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\]
The Cantor set is the set of numbers that remains after the inductive process of removing the middle thirds of the intervals in the previous step.  It is not obvious what is left (if anything), but consider the fact that as we discard an open interval representing the middle third of each closed interval, the endpoints of the original interval are always retained.  So $0$ and $1$ belong to $C_0$ and carry forward to $C_1$.  They are also endpoints of intervals in $C_1$, so they carry forward to $C_2$, and so on, so we can say that at the very least, $C$ contains every one of the endpoints of the closed intervals we produced during its construction.  Therefore, not only is the Cantor set not empty, but it is at least countably infinite.
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Now consider whether the Cantor set contains anything else.  We started with the interval $[0,1]$ which has length $1$, and discarded one interval of length $1/3$, two intervals of length $1/9$, four intervals of length $1/27$, and,  
in general,
\[
2^{n-1}\quad\mbox{intervals of length}\quad \frac{1}{3^n}
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\]
The combined lengths of the discarded intervals is
\[
L=\sum_{n=1}^\infty\frac{2^{n-1}}{3^n}=\frac{1}{3}\sum_{n=0}^\infty\left(\frac{2}{3}\right)^n=\frac{1}{3}\left(\frac{1}{1-\frac{2}{3}}\right)=1
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\] 
So the total length of the discarded intervals is one, meaning that the Cantor set has zero length.  All of this suggests a rather sparse collection consisting of the interval endpoints and not much else. 
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However, we can present an argument that $C$ is actually uncountable.  Suppose we construct an index $a_n$ for each element $x\in C$ as follows: Since $C$ is the intersection of all of the $C_i$, any point in $C$ belongs to all of them. Starting with $C_1$, define 
\[
a_1=\left\{\begin{array}{lcl}
0 &\mbox{if}& x\mbox{ is in the left half of } C_1\\
1 &\mbox{if}& x\mbox{ is in the right half of } C_1\\
\end{array}\right.
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\]
To create $C_2$, each half of $C_1$ is split into thirds, and the open middle third discarded.  Consequently, 
if $x$ belongs to, say, the left half of $C_1$, then when we form $C_2$ it must belong to either the left or the right half \textit{of the left half } of $C_1$.  So again we have two choices, and we will define, in this case,
\[
a_2=\left\{\begin{array}{lcl}
0 &\mbox{if}& x\mbox{ is in the left half of the left half of} C_1\\
1 &\mbox{if}& x\mbox{ is in the right half of the left half} C_1\\
\end{array}\right.
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\]
We continue in this fashion to define $a_3, a_4, a_5 \ldots$ either zero or one depending on whether $x$ lies in the left or right half of its previous interval when that interval is divided.  As a result, we end up with a sequence $a_n$ of zeros and ones for each point in $C$.  Because the endpoints are always retained, we know that none of the closed intervals we generate are empty, so there should be at least one point of $C$ belonging to every closed interval in the sequence of nested intervals designated by the sequence of zeros and ones $a_n$.  
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So, we expect there are at least as many points in $C$ as there are infinite sequences of zeros and ones.  However, 
the number of sequences of zeros and ones is uncountable.  To prove this, consider Cantor's diagonalization argument (Remark 1.39 on page 36 in the text), with the decimal sequences replaced by binary sequences.  We construct a sequence not in the list by taking the $k^{th}$ digit of the sequence under construction to be different from the $k^{th}$ digit of the $k^{th}$ sequence in the list. 
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So we have the remarkable conclusion that although $C$ has length $0$, it is uncountable.  
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In addition, note that since the endpoints of $C_0$, zero and one, are rational, all of the other endpoints are rational, which means that the set of endpoints is a subset of the rationals $\mbox{Q}$, and therefore is \textit{at most countable}.  So, the vast majority of the points in $C$ belong to $C$ minus the set of endpoints or $C\setminus\{\mbox{endpoints}\}$, which is uncountable.
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Finally, we add that although $C$ contains far more than the endpoints of the intervals, \textit{it does not contain any intervals}. To show this, suppose some interval $I=[a,b]\subseteq C$ where $a,b\in[0,1]$ and $a<b$.  Then every point in $I$ belongs to every one of the $C_n$, which means $I\subseteq C_n$ for all $n\in\mathbb{N}$.  Suppose $\epsilon>0$ is the length of  $I=b-a$.  The length of the intervals in $C_n$ is $1/3^n$, so by taking $n$ large enough that
\[
n > \log_3\left(\frac{1}{\epsilon}\right)
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\]
we have that $C_n$ is shorter than $I$, which it supposedly contains.  The contradiction shows that $C$ does not contain any intervals.
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