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\lecture{notes13}{Lecture Notes}{16 Oct 97}

The topic of this lecture is infinite sets.

Infinite sets can be tricky to think about.  For example, last time we
saw the Pigeonhole Principle, which says that if a big set is mapped
to a small set, then some two items of the big set must be mapped to
the same item in the small set.  We also saw that the Pigeonhole
Principle appears to break down for infinite sets.  In particular, the
``big'' set $\{ 0, 1, 2, 3, 4, \ldots \}$ of all natural numbers is
mapped one-to-one to the ``small'' set $\{ 0, 2, 4, 6, 8, \ldots
\}$ of even numbers by the function $f(n) = 2n$.  This raises confusion about
what is meant by ``big'' and ``small'' infinite sets.  One of the key
points of this lecture is sorting out the relative ``bigness'' of
various infinite sets.

This subject is more abstract than others in 6.042.  Nevertheless, the
techniques we develop for dealing with infinite sets will be quite
useful later on in 6.045 when you learn about computability.

\section{Finite Sets vs. Infinite Sets}

Let's start by distinguishing finite sets from infinite sets.

\begin{defin*}
A set is {\em finite} if there exists an $n \in {\cal N}$ such that
$|S| = n$.
\end{defin*}

\begin{defin*}
A set is {\em infinite} if for all $n \in {\cal N}$, $|S| \geq n$.
\end{defin*}

These definitions imply that every set is either finite or infinite.
For example, for all $n$, the set $\{ 0, 1, 2, 3, \ldots, n \}$ is
finite.  On the other hand, the set of all natural numbers, ${\cal N}
= \{ 0, 1, 2, 3, \ldots \}$, is infinite.  Here is a more interesting
example: 

\begin{theorem}
The set $P$ of all primes is infinite.
\end{theorem}

\begin{proof}
The proof is by contradiction.  Assume for the purpose of
contradiction that $P$ is finite.  By the definition of a finite set,
there exists an $n \in {\cal N}$ such that $|P| = n$.  Denote the
elements of $P$ by $p_1, p_2, \ldots, p_n$.  Now let $m = p_1 p_2
\cdots p_n + 1$.  Since $m > p_i$ for all $i$, $m$ cannot be a prime.
Hence $m$ must be divisible by some prime $p_i$:

\begin{eqnarray*}
& & p_i \mid m \\ & \Rightarrow & p_i \mid p_1 p_2 \cdots p_n + 1 \\ &
\Rightarrow & p_i \mid 1
\end{eqnarray*}

The first implication follows by substituting the definition of $m$.
The second follows because $p_i$ divides the product $p_1 p_2 \cdots
p_n$, and so must divide 1 in order to divide the sum.  But no prime
divides 1, so we have a contradiction.  Therefore, there are an
infinite number of primes.
\end{proof}

Proving that the set of primes is infinite is relatively easy, but the
next example shows that determining whether a set is finite or
infinite can be tricky.

\begin{defin*}
A {\em twin prime} is a pair $(n, n+2)$ such that both $n$ and $n + 2$
are prime.
\end{defin*}

Examples of twin primes include $(3, 5)$, $(5, 7)$, $(11, 13)$, $(17,
19)$, and $(101, 103)$.  Is the set of twin primes finite or infinite?
If you settle this question, then you can probably wheedle a few extra
credit points from your TA.  Mathematicians have tried unsuccessfully
to solve this problem for more than {\em two thousand years}!

\section{Countable Sets}

A nice property of finite sets is that we can write out their elements
in a list.  For example, we can write out the set of integers between
1 and n as $\{ 1, 2, 3, \ldots, n \}$.  We can also express some
infinite sets in a list format:

\begin{eqnarray*}
{\cal N} & = & \{ 0, 1, 2, 3, \ldots \} \\
\mbox{even numbers} & = & \{ 0, 2, 4, 6, \ldots \} \\
\mbox{primes} & = & \{ 2, 3, 5, 7, 11, \ldots \}
\end{eqnarray*}

Sets whose elements can be listed are called {\em countable sets}.
The reason for the name is that once we list the elements, there is a
first element, a second element, etc.  We can then count through the
items one after another so that every element of the set is eventually
counted.  Countable sets are formally defined in terms of bijections
with the natural numbers.

\begin{defin*}
An infinite set $S$ is said to be {\em countably infinite} iff there
exists a bijection $f: {\cal N} \mapsto S$.  A set is {\em countable}
if it is finite or countably infinite.
\end{defin*}

Many familiar sets are countable: ${\cal N}$, the even numbers,
primes, integers modulo a constant $k$, etc.

The formal definition of a countable set is equivalent to the notion
of a listable set, since listing the elements of a set gives a
bijection with ${\cal N}$ and vice-versa.  If we can list the elements
of $S$ as $s_0, s_1, s_2,
\ldots$, then we can construct a bijection $f: {\cal N} \mapsto S$
defined by $f(i) = s_i$.  In the reverse direction, if we have such a
bijection, then $f(0), f(1), f(2), \ldots$ is a list of all the
elements of $S$.

Sometimes we can list the elements of a set $S$, but the corresponding
bijection $f: {\cal N} \mapsto S$ is very hard to compute.  For
example, in principle we could list the elements in any subset of
${\cal N}$.  However, most of these subsets are bizarre and difficult
to describe; writing a program to compute $f(i)$ might be difficult or
even impossible.  The good news is that to prove that $S$ is
countable, we must only prove that a bijection $f: {\cal N} \mapsto S$
exists; we do not have to write $f$ explicitly or to give an algorithm
to compute $f(i)$.

\section{Uncountable Sets}

There would be little need for intuition and definitions about
countable sets if all sets were countable; unfortunately, many
important sets are not countable.  Sets that are not countable are
called {\em uncountable}.

\begin{theorem*}
The set ${\cal R}$ of real numbers is uncountable.
\end{theorem*}

This theorem says that there is no way to make a list of all the real
numbers.  Naturally, one is immediately tempted to try anyway:

\[
0,\ 1,\ 0.5,\ 0.75,\ 0.1,\ 0.05,\ 0.001,\ \ldots
\]

This is not going to be so easy!  For example, where are we going to
stick in $\sqrt{2}$ and $\pi^{-173}$?  Later we will see clever ways
of listing the elements of some seemingly ``big'' sets like the
rational numbers, but for the real numbers we can actually prove that
no such listing exists.

We'll first work through the proof informally, and then write it up.
The argument is by contradiction; we assume for the purpose of
contradiction that ${\cal R}$ is countable.  This means that there
exists a bijection $f: {\cal N} \mapsto {\cal R}$.  We can then make a
list of all the real numbers that looks something like the one shown
below.  (The extra spaces only serve to highlight a diagonal sequence
of digits discussed below.)

\begin{verbatim}
                    f(0) =  76. 5 430000000...
                    f(1) = 163.7 7 77777777...
                    f(2) =   2.78 7 8787878...
                    f(3) =  -1.112 1 231234...
                    f(4) =   3.1415 9 26535...
                    f(5) = -15.37273 2 3928...
                          ...etc...
\end{verbatim}

The plan is to reach a contradiction by constructing a real number
that appears nowhere in the list; this will prove that $f$ is not
surjective and contradict the assumption that $f$ is a bijection.

Consider the number defined by the diagonal digits in the above table,
$0.577192\ldots$.  Now suppose that we add 5 to every digit, working
modulo 10.  This gives $0.022647\ldots$; call this value $r$.  This
quantity $r$ is totally bizarre, but it is a well-defined real number,
since we have specified all the digits.

By our construction of $r$, $f(0)$ differs from $r$ in the first
position to the right of the decimal.  Similarly, $f(1)$ differs from
$r$ in the second decimal position, $f(2)$ differs in the third
position, and so forth.  In general, $f(i)$ differs from $r$ in the
$(i+1)$-st position.  So for all $i$, $f(i)$ is not equal to $r$.
This implies that the real number $r$ does not appear anywhere in our
list!

There is one irritating problem in the preceding paragraph that must
be fixed.  Some numbers have two decimal representations.  For
example, $1 = 0.999\ldots$.  Therefore, distinct decimal
representations do not necessarily correspond to distinct numbers.
Even though $r$ disagrees with every $f(i)$ in at least one digit, the
two might be equal.  However, our construction ensures that $r$
actually differs from $f(i)$ by at least $4 \cdot 10^{-(i+1)}$.  For
example, $1.0002999\ldots$ differs from $1.0007000\ldots$ by $4 \cdot
10^{-4}$.

Here is how the same argument might be written as a formal proof.

\begin{proof}
We must show that there is no bijection $f: {\cal N} \mapsto {\cal
R}$.  The proof is by contradiction.  Assume for the purpose of
contradiction that such a bijection $f$ exists.  Let $r$ be the real
number with decimal expansion $0.r_0r_1r_2r_3\ldots$ where $r_i$ is $5
\ +$ the $(i+1)$-st decimal digit of $f(i)$, taken modulo 10.  
For all $i$, $f(i) \neq r$, since the $(i+1)$-st digits of $f(i)$ and
$r$ disagree by 5 and so $|r - f(i)| \geq 4 \cdot 10^{-(i+1)}$.  This
implies that $f$ is not surjective, contradicting the assumption that
$f$ is a bijection.
\end{proof}

This is called a ``diagonalization argument'', because of the
construction based on diagonal entries.  Diagonalization arguments are
used all the time to prove that there is no bijection between two
infinite sets.  The generic idea is to assume for the purpose of
contradiction that a bijection $f$ exists.  Then an element $r$ of the
range is constructed that differs a little bit from $f(\alpha)$ for
every $\alpha$ in the domain.  (In this case, for example, the real
number $r$ differs from $f(k)$ in the $(k+1)$-st decimal position for
every natural $k$.)  The existence of such an $r$ shows that $f$ is
not surjective, contradicting the assumption that $f$ is a bijection.
We conclude that no bijection between the two sets exists.

Diagonalization arguments are important in logic and computer science.
For example, this type of argument is used to show that there are some
problems that can not be solved by a computer no matter how much time
you give them.  A similar method is used in logic to prove G\"{o}del's
Theorem, which says that there are some true theorems for which no
proof exists!

\section{Countable or Uncountable?}

We now know that the natural numbers are countable and the real
numbers are uncountable.  In general, the first question about an
infinite set is whether it is countable or uncountable.  Let's answer
this question for some simple sets.

\subsection*{The Integers}

\begin{theorem}
The set ${\cal Z}$ of integers is countable.
\end{theorem}

\begin{proof}
To prove that that the integers are countable, we must prove that
there exists a bijection $f: {\cal N} \mapsto {\cal Z}$.  The integers
can be listed as follows: 0, 1, -1, 2, -2, 3, -3, 4, -4, etc.  This
listing defines the desired bijection, e.g. $f(0) = 0$, $f(1) = 1$,
$f(2) = -1$, $f(3) = 2$, etc.
\end{proof}

This is the typical way to prove that a set is countable--- just show
that you can list all of the elements.  This implicitly defines a
bijection with ${\cal N}$.  As noted above, we only have to prove that
a bijection $f$ exists; $f$ need not be a function that we can
actually compute.  In this case, however, $f$ can be written simply:

\[
f = \left \{
\begin{array}{ll}
- \frac{i}{2} & i \mbox{ even} \\
\frac{i+1}{2} & i \mbox{ odd} \\
\end{array}
\right.
\]

The method of proving a set countable by listing elements has one
potential pitfall.  As an example, the following listing of the
integers is incorrect: $0, -1, -2, -3, \ldots, 1, 2, 3, \ldots$.  The
problem is that the ``$\ldots$'' in the middle of the list represents
a infinite number of elements.  As a result, we never actually get
around to listing the positive integers; in particular, there is no
$i$ such that $f(i) = 1$.  This means that the function $f$ defined by
this listing is not a bijection.

\subsection*{The Half Integers}

Define the set $Z^{1/2}$ as follows:

\begin{eqnarray*}
Z^{1/2} & = & \{ \frac{x}{2} \mid x \in {\cal Z} \} \\
        & = & \{ 0, 1, -1, 2, -2, \ldots \} \cup
              \left\{ \frac{1}{2}, -\frac{1}{2}, \frac{3}{2}, -\frac{3}{2},
		\ldots \right\}
\end{eqnarray*}

In other words, $Z^{1/2}$ consists of all integers and all half
integers.  This set is countable.  We have written $Z^{1/2}$ above as
the union of two countable sets.  We can list all the elements of
$Z^{1/2}$ by interleaving these two lists:

\[
Z^{1/2} = \left\{ 0, \frac{1}{2}, 1, -\frac{1}{2}, -1,
	\frac{3}{2}, 2, -\frac{3}{2}, -2, \ldots \right\}
\]

This trick generalizes to a theorem.

\begin{theorem}
\label{th:union}
If $A$ and $B$ are countable, then $A \cup B$ is countable.
\end{theorem}

\begin{proof}
Since $A$ and $B$ are countable, we can list the elements:

\begin{eqnarray*}
A & = & \{ a_1, a_2, a_3, \ldots \} \\
B & = & \{ b_1, b_2, b_3, \ldots \} \\
\end{eqnarray*}

Therefore, we can list the elements of the union as follows:

\[
A \cup B  =  \{ a_1, b_1, a_2, b_2, a_3, b_3, \ldots \}
\]

There are two complications.  First, sets $A$ and $B$ may contain
common elements, so an element should be omitted from the list if it
has already appeared.  Second, either or both sets may be finite.  If
at some point all elements of one set have been listed, continue
listing elements of the other set without interleaving.
\end{proof}

\subsection*{The Rationals}

Recall that the set ${\cal Q}$ of rational numbers consists of all
ratios of integers with non-zero denominator:

\[
{\cal Q} = \{ a/b \mid a, b \in {\cal Z}, b \neq 0 \}
\]

There are a lot of rational numbers!  It is easy to imagine that we
could not list all the rational numbers just as we could not list all
the real numbers.  In fact, however, there is a clever way to list all
the rationals, and this proves that the set ${\cal Q}$ is countable.

\begin{theorem}
\label{th:rationals}
The set ${\cal Q}$ of rational numbers is countable.
\end{theorem}

\begin{proof} {\em (Failed attempt.) }
To prove that that the rational numbers are countable, we must prove
that there exists a bijection $f: {\cal N} \mapsto {\cal Q}$.  The
rational numbers can be listed as follows:

\[
0,
\frac{1}{1}, \frac{2}{1}, \frac{3}{1}, \ldots,
\frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \ldots,
\frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \ldots
\]

Omit from the listing all numbers that have already appeared.  (For
example, $2/2$ and $3/3$ should be omitted, since the rational number
1 already appeared as $1/1$.)  This listing implicitly defines the
desired bijection $f$.
\end{proof}

There are two problems with this proof, one minor and one serious.
The minor problem is that we completely forgot the negative rational
numbers.  But the negative rationals are easy to add; we will show how
in the correct proof given below.  The killer problem is that we used
``$\ldots$'' in the middle of the list to represent an infinite number
of elements.  This means that we will never reach, say, 1/2; that is,
there is no $i$ such that $f(i) = 1/2$.  This means that $f$ is not a
bijection as claimed.  We need to list the rational numbers in a
completely different order to make the proof work.

\begin{proof} {\em (Successful attempt.) }
We can list the positive rational numbers in the order indicated in
Figure~\ref{fig:rationals}, omitting numbers that have already
appeared in the list.  This proves that the positive rationals are
countable.

\begin{figure}[htbp]
\centerline{\psfig{figure=rationals.eps,height=2.5in}}
\caption{\em This digram shows the ordering of rationals used in the proof
of Theorem~\ref{th:rationals}.  This ordering technique is called
``dovetailing''.}
\label{fig:rationals}
\end{figure}

Negating each number in the figure establishes that the negative
rationals are countable.  The set of all rationals is the union of
positive rationals, negative rationals, and $\{ 0 \}$.  Since these
three sets are countable, two applications of Theorem~\ref{th:union}
show that the set of all rationals is countable.
\end{proof}

The ordering technique used in the proof and indicated in
Figure~\ref{fig:rationals} is called ``dovetailing''.  This technique
is often used in 6.045.  We can use dovetailing to prove that the
product of any two countable sets is countable:

\begin{theorem}
If $A$ and $B$ are countable, then $A \times B$ is countable.
\end{theorem}

\begin{proof}
Since $A$ and $B$ are countable, we can list the elements of $A$ as
$\{a_1, a_2, a_3, \ldots\}$ and the elements of $B$ as $\{b_1, b_2, b_3,
\ldots\}$.  We can obtain a listing of $A \times B$ by dovetailing:

\begin{eqnarray*}
&& (a_1, b_1), \\
&& (a_1, b_2), (a_2, b_1), \\
&& (a_1, b_3), (a_2, b_2), (a_3, b_1), \\
&& (a_1, b_4), (a_2, b_3), (a_3, b_2), (a_4, b_1), \\
&& \quad \quad \ldots \quad \quad \ldots
\end{eqnarray*}

If either or both sets $A$ and $B$ are finite, omit from the listing
all pairs containing non-existent elements $a_i$ or $b_j$.
\end{proof}

Note that the ordering of elements in this proof corresponds to the
ordering in Figure~\ref{fig:rationals}.  In fact, the proof that the
rationals are countable used the idea of the above proof with $A = B =
{\cal Z}^+$.

\subsection*{A Paradox}

Here is a set that appears to be both countable and uncountable!  This
is impossible, so one argument must be incorrect, but which one?

Define $S_n$ to be the set of $n$-digit decimal numbers.  For example:

\begin{eqnarray*}
S_1 = \{ .0, .1, .2, \ldots, .9 \} \\
S_2 = \{ .00, .01, .02, \ldots, .99 \} \\
\end{eqnarray*}

Now let $S = \bigcup_{n=1}^\infty S_n$.  Is $S$ countable?

{\em Argument that $S$ is uncountable: } We will prove that $S$ is
uncountable by the same diagonalization argument that we used to prove
that the set ${\cal R}$ of real numbers is uncountable.  Assume for
the purpose of contradiction that we can list the elements of $S$ in
some order $s_0, s_1, s_2, \ldots$.  Define $r = .r_0r_1r_2\ldots$
where $r_i$ is $5\ +$ the $(i+1)$-st decimal digit of $s_i$ taken
modulo 10.  Since $S$ contains decimal numbers of every length, $r$
must be in $S$.  However, $r$ differs from every element of $S$ by at
least one digit.  Therefore, $r$ is not in $S$, and we have a
contradiction.

{\em Argument that $S$ is countable: } We can list the elements of $S$
by listing the elements of $S_1$, followed by $S_2$, followed by
$S_3$, etc.  Since each of these sets is finite, every set is fully
listed eventually.

The first argument is incorrect, and the second argument is correct.

The false statement in the first argument is, ``Since $S$ contains
decimal numbers of every length, $r$ must be in $S$.''  The set $S$
does contain decimal numbers of every {\em finite} length, but no
numbers with an infinite number of non-zero digits to the right of the
decimal.  For example, the number $1/9 = 0.11111\ldots$ is not
contained in any set $S_i$ and therefore is not contained in the union
$S$.  As a result, there are many real numbers not in $S$.  For this
reason, the statement, ``$r$ must be in $S$'' is false.  This
eliminates the contradiction from the proof by contradiction!

The second argument is correct; the union of a countable number of
finite sets is always countable.  In fact, for homework you will show
that a union of a countable number of countable sets is always
countable.

The distinction between countable and uncountable sets is so important
that it distinguishes entire fields of mathematics.  Discrete math is
the study of countable sets; calculus, analysis, and continuous math
all study uncountable sets.  In this course, we will stick to discrete
mathematics after today; all sets will be countable.

\section{Other Infinite Sets}

Two sets are considered to be the same size if there is a bijection
between them.  For example, the set of natural numbers is the same
size as the set of rational numbers because there is a bijection $f:
{\cal N} \mapsto {\cal Q}$.  On the other hand, the set of natural
numbers is not the same size as the set of real numbers because we
proved there is no bijection between these sets.

More generally, we say that:

\begin{itemize}
\item $|A| \leq |B|$ iff there exists an injection $f: A \mapsto B$
\item $|A| \geq |B|$ iff there exists a surjection $f: A \mapsto B$
\item $|A| = |B|$ iff there exists a bijection $f: A \mapsto B$
\item $|A| < |B|$ iff $|A| \leq |B|$ and $|A| \neq |B|$
\item $|A| > |B|$ iff $|A| \geq |B|$ and $|A| \neq |B|$
\end{itemize}

For example, $|{\cal N}| < |{\cal R}|$.  This fact follows from
$|{\cal N}| \leq |{\cal R}|$ and $|{\cal N}| \neq |{\cal R}|$.  The
first statement holds because there is an injection from ${\cal N}$ to
${\cal R}$; namely, each natural number is mapped to itself.  The
second statement holds because we have proven that there is no
bijection between ${\cal N}$ and ${\cal R}$.

In this sense, ${\cal N}$ and ${\cal R}$ are two different kinds of
infinities.  In fact, there are many types of infinities, and there is
a whole branch of mathematics to deal with them.  The most common
types of infinities are based on power sets.  (For now, we will denote
the power set of $S$ by ${\cal P}(S)$.)  These types of infinity are
denoted by the Hebrew letter aleph, $\aleph$, with a subscript.  A
bigger subscript implies a bigger type of infinity.  Here are the
first few:

\begin{eqnarray*}
\aleph_0 & = & |{\cal N}| \\
\aleph_1 & = & |{\cal P}({\cal N})| = |{\cal R}| \\
\aleph_2 & = & |{\cal P}({\cal P}({\cal N}))| = |{\cal P}({\cal R})| \\
	& \ldots
\end{eqnarray*}

Each new type of infinity is obtained by taking the power set of the
preceding type of infinity.  (We will not prove the assertion on the
second line that the set of real numbers is the same size as the power
set of the natural numbers.)  It is not immediately clear, however,
that sets of size $\aleph_2$ are really any larger than sets of size
$\aleph_1$, or more generally that taking the power set of $S$ gives a
larger set than $S$ itself.  This fact can be proven, however.

\begin{theorem}
For all sets $S$, $|S| < |{\cal P}(S)|$.
\end{theorem}

This implies that $\aleph_0 < \aleph_1 < \aleph_2 < \ldots$, and so
there is a whole infinite hierarchy of infinite sets!  The proof of
the theorem is really convoluted; the main idea is drawn from
``Russel's Paradox''.

\begin{proof}
We have to prove that $|S| \leq |{\cal P}(S)|$ and $|S| \neq |{\cal
P}(S)|$.

{\em $|S| \leq |{\cal P}(S)|$: } We must show that there exists an
injection $f: S \mapsto {\cal P}(S)$.  The function $f(\alpha) = \{
\alpha \}$ for all $\alpha \in S$ is just such an injection.

{\em $|S| \neq |{\cal P}(S)|$: } The proof is by contradiction.
Assume for the purpose of contradiction that $|S| = |{\cal P}(S)|$.
Then there is a surjection $f: S \mapsto {\cal P}(S)$.  (In fact,
there is a bijection, but we do need this fact.)  To show a
contradiction, we will define an item in ${\cal P}(S)$ that is not
covered by $f$; this will contradict the assertion that $f$ is a
surjection.  In particular, let $R = \{ a \mid a \in S, a \not\in f(a)
\}$.  The set $R$ is a subset of $S$, and so $R$ is an element of the
power set of $S$.  Since $f$ is a surjection, there must be an $a
\in S$ such that $f(a) = R$.  Now there are two cases: either $a \in
R$ or $a \not\in R$.

{\em ($a \in R$): } By the definition of the set $R$, this implies
that $a \not\in f(a) = R$, a contradiction.

{\em ($a \not\in R$): } Now the definition of the set $R$ implies that
$a \in f(a) = R$, another contradiction.

The contradictions in both cases imply that our assumption was wrong;
therefore, $|S| \neq |{\cal P}(S)|$ and so $|S| < |{\cal P}(S)|$.
\end{proof}

We can construct many types of infinity using power sets, but what
about completely different types of infinity?  For example, is there
an infinite set $S$ that is larger than the natural numbers, but
smaller than the real numbers?  This question was posed by Cantor in
the 1870's and was not resolved for almost a century.

In 1900 David Hilbert posed 24 unsolved problems for the new century
and this was first on his list.  Hilbert conjectured that no such set
existed, and this conjecture became known as the Continuum
Hypothesis.

In 1963, Paul Cohen answered the question.  He did not prove that such
a set existed or did not exist, but rather showed that {\em both}
scenarios are consistent with the axioms of set theory.  As a
consequence, if the axioms of set theory are consistent, then there
is no way to prove the Continuum Hypothesis true or false!
\end{document}
