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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Discussion 2 -- The Cantor Set}
\centerline{October 24-26, 2005}
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\noindent
{\it Background}
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In this discussion, we will study some of the surprising properties
of a subset of the real number system called the {\it Cantor set}.
This was first constructed by the same mathematician Georg Cantor
who we met earlier when we discussed the uncountability of ${\bf R}$.
The Cantor set is produced by the following process:
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\item{1)}  Start with the closed interval $C_0 = [0,1]$.
\item{2)}  Discard the open ``middle third'' $(1/3,2/3)$
interval to make 
$$C_1 = [0,1/3] \cup [2/3,1].$$
\item{3)}  Discard the open ``middle third'' of each half 
of $C_1$ to make 
to make 
$$C_2 = [0,1/9] \cup [2/9,1/3] \cup [2/3,7/9] \cup [8/9,1].$$
\item{4)}  Discard the open ``middle third'' of each of the four 
pieces of $C_2$ to make $C_3$, and so on.
\item{5)}  Repeat this process of removing the middle thirds 
to produce $C_n$ for all $n\ge 1$.  
\item{6)}  The Cantor set $C$ is then 
$$C = \bigcap_{n=1}^\infty C_n.$$
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\noindent
{\it Discussion Questions}
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\item{A)}  Show that the sum of the lengths of all of the intervals 
{\it removed} in the process described above is 1.  (Note: there are 
infinitely many of them, so this is an infinite series!)
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\item{B)}  Since the whole interval $[0,1]$ has length 1, 
you might think that means that $C = \emptyset$. 
But in fact that is {\it far from true}.  In fact, $C$ is still
an {\it uncountably infinite} set of real numbers.  Here is one
way to see this:
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\itemitem{1)} We usually use base 10 expansions to represent
real numbers, but any other integer base $b > 1$ would work
just as well.  Let's consider base $b = 3$ or ``ternary'' expansions.
The ternary expansion of a whole number like $n = 24$ would be found like
this:  $24 = 2\cdot 3^2 + 2\cdot 3 + 0\cdot 1$.  So we would 
say $24 = (220)_3$.  If a number has a fractional part, then 
that is represented by negative powers of $3$.  
The ternary digits of a number $x$ in $[0,1]$ are integers $d_i = 0,1$, or $2$
appearing in a series
$$x = {d_1\over 3} + {d_2\over 3^2} + {d_3\over 3^3} + \cdots $$
Show that every such series (that is, for every possible 
choice of $d_i \in \{0,1,2\}$, $i \in {\bf N}$) converges to some real number in $[0,1]$.
\itemitem{2)}  What is true about the ternary digit $d_1$ for the numbers
removed in the middle half of $[0,1]$?  What is true about 
$d_1$ for the numbers in the Cantor set $C$?  Similarly, what is true about
the ternary digit $d_2$ for the number removed from $C_1$ to get $C_2$?
What is true about $d_2$ for the numbers in $C$? 
\itemitem{3)}  Let's consider the set of all sequences $(x_n)$ where
$x_n = 0$ or $2$ for each $n \in {\bf N}$.  Show that this 
set of sequences is an uncountable set by adapting the Cantor
Diagonalization proof we used to show $[0,1]$ is uncountable.  
\itemitem{4)}  Put together parts 1,2,3 to explain why $C$ is 
uncountably infinite.
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\item{C)}  Here is another surprising property of $C$.  We claim
that every number in the closed interval $[0,2]$ can be obtained
as $x + y$ for some $x,y \in C$.  (This is true even though $C$
actually contains {\it no intervals} itself!).  Follow this plan:
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\itemitem{1)}  First show by induction that for each $n\ge 1$,
every number in $[0,2]$ can be obtained as $x_n + y_n$ for some 
$x_n, y_n \in C_n$.  
\itemitem{2)}  The $(x_n)$ and $(y_n)$ are bounded sequences.
Deduce the desired statement by applying Bolzano-Weierstrass (and
other results we have seen, as needed!)

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\noindent
{\it Assignment}
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Writeups due Friday, October 28.
\bye
