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......@@ -10,6 +10,25 @@ correspondence from one to the other.
We write $A\sim B$.
\end{df}
\begin{problem} \label{ex:EqCardOfSomePairs}
For each, prove the two have the same cardinality.
\begin{items}
\item $\N$ and $\mathcal{E}=\set{2k\suchthat k\in\N}$,
\item $\open{-\pi/2}{\pi/2}$ and $\R$,
\item $\open{0}{1}$ and $\R$
\end{items}
\begin{answer}
\begin{items}
\item The function $\map{f}{\N}{\mathcal{E}}$ given by $f(n)=2n$ is
one-to-one and onto.
\item The function $\map{\tan}{\open{-pi/2}{\pi/2}}{\R}$ is one-to-one
and onto.
\item Consider the map $\map{t}{\open{0}{1}}{\R}$ given by
$t(x)=\tan(\pi(x-(1/2)))$.
\end{items}
\end{answer}
\end{problem}
\begin{problem} \label{ex:EquinumeruousIsEquivalence}
Prove that the relation~$\sim$ is an equivalence.
\begin{answer}
......@@ -43,7 +62,7 @@ A set is \definend{countable} if it is either finite or denumerable.
\end{df}
\begin{problem} \label{ex:IntegersCountable}
Prove.
For each, prove the two sets have the same cardinality.
\begin{exes}
\begin{exercise}
The set of integers is countable.
......@@ -61,33 +80,31 @@ Prove.
\end{exercise}
\begin{answer}
As in the prior item we will define a map
$\map{f}{\N}{\N\times\N}$ and only note that it is a correspondence,
without going through the verification details.
The function alternates between columns:
$0\mapsunder{} (0,0)$, $1\mapsunder{}(0,1)$, $2\mapsunder{}(1,0)$,
$3\mapsunder{}(0,2)$,
$4\mapsunder{}(1,1)$, $5\mapsunder{}(2,0)$, $6\mapsunder{}(0,3)$, \ldots.
$\map{f}{\N}{\N\times\N}$.
The function alternates between columns and rows.
It first covers the outputs $\sequence{x,y}$ where $x+y=0$, then those
where $x+y=1$, etc.
Here are the first few values.
\begin{center}
\begin{tabular}{r|cccccccccc}
\textit{Input} &$0$ &$1$ &$2$ &$3$
\textit{Input $n$} &$0$ &$1$ &$2$ &$3$
&$4$ &$5$
&$6$ &$7$ &$8$ &$\cdots$ \\ \cline{2-11}
\rule{0em}{2.25ex}
\textit{Output} &$\seq{0,0}$ &$\seq{0,1}$ &$\seq{1,0}$ &$\seq{0,2}$
\textit{Output $f(n)$} &$\seq{0,0}$ &$\seq{0,1}$ &$\seq{1,0}$ &$\seq{0,2}$
&$\seq{1,1}$ &$\seq{2,0}$
&$\seq{0,3}$ &$\seq{1,2}$ &$\seq{2,1}$ &$\cdots$
\end{tabular}
\end{center}
This is clearly a correspondence.
\remark a sketch helps here.
\remark sketch helps.
The values associated with successive arguments trace out the
diagonals of the array~$\N\times\N$.
\begin{center}
\newcommand{\chanpoint}[1]{\makebox[0em]{$(#1)$}}
\newcommand{\chanarrow}{{\color{darkii} $\searrow$}}
\setlength{\unitlength}{0.5in}
\setlength{\unitlength}{0.45in}
\begin{picture}(4,4)
\put(0,0){\chanpoint{0,0}}
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......@@ -238,23 +255,34 @@ possible if $j_0=j_1$).
\begin{problem}
Prove that the set of rational numbers is countable.
\begin{answer}
We will show that the set on nonnegative rationals
We will argue that the set of nonnegative rationals
$Q=\setbuilder{q\in\Q}{q\geq 0}$ is countable.
With that set countable, a function that alternates sign like the one in
Exercise~\ref{ex:IntegersCountable} will show that~$\Q$ is countable.
With that, a function that alternates sign like the one in
Exercise~\ref{ex:IntegersCountable} will count the entire set~$\Q$.
Consider the function $\map{g}{\N\times\N}{\Q}$ where $g(p,q)$
is the rational number $(p+1)/(q+1)$.
It is not quite onto; every rational number is the image of some
$(p,q)$ except for $0$.
We will give a procedure to define the enumerating function
$\map{f}{\N}{\Q^+}$.
The prior exercise shows that the set $\N\times\N$ is countable;
let it be counted by~$\map{h}{\N}{\N\times\N}$.
Define $f(0)=0$.
For $f(1)$, enumerate $h(0)$, $h(1)$, \ldots $h(i_1)$ until there is a value
$\seq{x_1,y_1}$ such that $y_1\neq 0$ and the fraction $x_1/y_1$
is not equal to~$0$.
Such a value must appear because $h$ enumerates $\N\times\N$.
Set $f(1)=x_1/y_1$.
The Cartesian product $\N\times\N$ is denumerable so there is a
correspondence $\map{f}{\N}{\N\times\N}$.
Define a function $\map{h}{\N}{\Q}$ by $h(i)=0$ if $i=0$ and $h(i)=f(i-1)$
if $i>0$.
The range is $\range(h)=\range(g)\union\set{0}$ so~$h$ is onto.
By Exercise~\ref{ex:OntoFromNatsToSetEquivToCountable} the
rationals~$\Q$ form a countable set.
Iterate.
Step~$s+1$ starts with a list $f(0)$, $f(1)$, \ldots{}
$f(s)$,
such that $f(s)=x_s/y_s$ where $h(i_s)=\seq{x_s,y_s}$.
Produce $f(s+1)$ by enumerating $h(i_s+1)$, $h(i_s+2)$, \ldots{}
until there is a value
$h(i_{s+1})=\seq{x_{s+1},y_{s+1}}$ such that $y_{s+1}\neq 0$, and the
fraction $x_{s+1}/y_{s+1}$ does not appear in the list
$f(0)$, $f(1)$, \ldots $f(s)$.
Such a value must appear because $h$ enumerates $\N\times\N$.
Set $f(s+1)=x_{s+1}/y_{s+1}$.
\end{answer}
\end{problem}
......@@ -264,38 +292,44 @@ possible if $j_0=j_1$).
$\powerset(\N)$
\end{exercise}
\begin{answer}
We will show that no $\map{f}{\N}{\powerset(\N)}$ is onto.
We will prove that no function $\map{f}{\powerset(\N)}{\N}$ is onto.
Consider $R=\setbuilder{n\in\N}{n\notin f(n)}$.
Clearly $R\subseteq \N$ and so $R\in\powerset(\N)$.
Clearly $R\in\powerset(\N)$.
We will show that there is no $i\in\N$ such that $R=f(i)$.
For contradiction assume that $R=f(i)$.
Consider whether $i\in R$.
If $R=f(i)$ and $i\in R$ then that contradicts the definition of $R$ as the
Consider whether $i$ is an element of~$R$.
Assuming $i\in R$ contradicts the definition of $R$ as the
collection of integers~$n$ such that $n\notin f(n)$.
The other $R=f(i)$ possibility is that $i\notin R$, which
is also a contradiction
Assuming $i\notin R$
also leads to a contradiction
because then $i$~satisfies the defining criteria for membership in~$R$, namely
that $i\notin R$, and therefore $i\in R$.
Either $R=f(i)$ case gives a contradiction and so $R\neq f(i)$.
Both of the two possibilities are impossible, and so $R\neq f(i)$.
\end{answer}
\begin{exercise}
$\R$
\end{exercise}
\begin{answer}
We will show that there is a one-to-one map~$\map{f}{\powerset(\N)}{\R}$.
If $\R$ were countable\Dash if there were a one-to-one
function $\map{g}{\R}{\N}$\Dash then
that would give a one-to-one function~$\composed{g}{f}$
from $\powerset(\N)$ to~$\N$,
contradicting the prior item.
We will show that there is a one-to-one
map~$\map{f}{\powerset(\N)}{\R}$.
That gives the result because
if $\R$ were countable\Dash if there were a correspondence
$\map{g}{\R}{\N}$\Dash then
that would give an one-to-one function~$\composed{g}{f}$
from $\powerset(\N)$ to $\N$,
contradicting the prior item and
Exercise~\ref{ex:OntoFromNatsToSetEquivToCountable} above.
We define the action of~$f$ on a set~$A\in\powerset(\N)$ by
giving an associated real number between $0$ and~$1$.
We will give this number's binary expansion, instead of its more usual
decimal expansion.
Let this number be such that its $i$-th binary place is $\charfcn{A}(i)$,
that is, the $i$-th binary place is $1$ if $i\in A$ and is $0$ otherwise.
An input to~$f$ is a set~$A\in\powerset(\N)$.
We define $f$'s action by
giving an associated real number output.
(This output is in the interval $\closed{0}{1}$ but
that is not relevant to this proof.)
Let the output number be such that its $i$-th decimal place is
$\charfcn{A}(i)$,
that is, the $i$-th decimal place is~$1$ if $i\in A$ and is $0$~otherwise.
This function is one-to-one because two unequal sets differ in an
element~$a\in\N$, and the two image numbers then differ in the $a$-th place.
\end{answer}
......
......@@ -114,8 +114,10 @@ then
asy -fpdf bean_fcn.asy
asy -fpdf bean_fcn_comp.asy
asy -fpdf bean_fcn_elets.asy
asy -fpdf bean_fcn_onto_counter.asy
asy -fpdf bean_fcn_otoo_counter.asy
asy -fpdf bean_partition.asy
asy -fpdf bean_set.asy
# asy -fpdf bean_set.asy
asy -fpdf venn_comp.asy
asy -fpdf venn_comp_comp.asy
asy -fpdf venn_diff.asy
......
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