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Commits (1)
......@@ -5,20 +5,9 @@ that if two finite sets correspond then they have the same number of elements.
\begin{df}
Two sets have the \definend{same cardinality}
(or are \definend{equinumerous}) if there is a
or are \definend{equinumerous} if there is a
correspondence from one to the other.
We write $A\sim B$.
\end{df}
\begin{df}
A set is \definend{finite} if it has~$n$ elements for some $n\in\N$,
that is, if it has the same cardinality as
$\setbuilder{i\in\N}{i< n}=\set{0,1,\ldots,n-1}$.
Otherwise the set is \definend{infinite}.
% \end{df}\begin{df}
A set is \definend{denumerable} if it has the same cardinality
as $\N$.
A set is \definend{countable} if it is either finite or denumerable.
We write $A\sim B$.
\end{df}
\begin{problem} \label{ex:EquinumeruousIsEquivalence}
......@@ -42,6 +31,17 @@ Prove that the relation~$\sim$ is an equivalence.
\end{answer}
\end{problem}
\begin{df}
A set is \definend{finite} if it has~$n$ elements for some $n\in\N$,
that is, if it has the same cardinality as some initial segment
$\setbuilder{i\in\N}{i< n}=\set{0,1,\ldots,n-1}$ of the natural numbers.
Otherwise the set is \definend{infinite}.
% \end{df}\begin{df}
A set is \definend{denumerable} if it has the same cardinality
as $\N$.
A set is \definend{countable} if it is either finite or denumerable.
\end{df}
\begin{problem} \label{ex:IntegersCountable}
Prove.
\begin{exes}
......@@ -49,12 +49,12 @@ Prove.
The set of integers is countable.
\end{exercise}
\begin{answer}
We will define a map $\map{f}{\N}{\Z}$ and just note that it is clearly a
correspondence.
The map alternate between positive and negative values:
Consider the map $\map{f}{\N}{\Z}$
that alternates between positive and negative values:
$0\mapsunder{}0$, $1\mapsunder{}1$, $2\mapsunder{}-1$,
$3\mapsunder{}2$, $4\mapsunder{}-2$, $5\mapsunder{}3$, \ldots\,.
More formally, $f(n)=-n/2$ if $n$ is even and $f(n)=(n+1)/2$ if~$n$ is odd.
More formally, $f(n)=-n/2$ if $n$ is even and $f(n)=(n+1)/2$ if~$n$ is odd.
It is clearly a correspondence.
\end{answer}
\begin{exercise}
The set~$\N\times\N$ is countable.
......@@ -64,13 +64,23 @@ Prove.
$\map{f}{\N}{\N\times\N}$ and only note that it is a correspondence,
without going through the verification details.
The function alternates between columns in the same way that
Jackie Chan, when up against multiple bad guys, alternates among his
attackers.
Thus $0\mapsunder{} (0,0)$, $1\mapsunder{}(0,1)$, $2\mapsunder{}(1,0)$,
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.
\begin{center}
\begin{tabular}{r|cccccccccc}
\textit{Input} &$0$ &$1$ &$2$ &$3$
&$4$ &$5$
&$6$ &$7$ &$8$ &$\cdots$ \\ \cline{2-11}
\rule{0em}{2.25ex}
\textit{Output} &$\seq{0,0}$ &$\seq{0,1}$ &$\seq{1,0}$ &$\seq{0,2}$
&$\seq{1,1}$ &$\seq{2,0}$
&$\seq{0,3}$ &$\seq{1,2}$ &$\seq{2,1}$ &$\cdots$
\end{tabular}
\end{center}
\remark a sketch helps here.
The values associated with successive arguments trace out the
diagonals of the array~$\N\times\N$.
......@@ -170,8 +180,8 @@ Prove.
\begin{problem} \label{ex:OntoFromNatsToSetEquivToCountable}
Prove that the following are equivalent for a set~$A$:
(i)~$A$ is countable
(ii)~$A$ is empty or there is an onto function from $\N$ to~$A$
(i)~$A$ is countable,
(ii)~$A$ is empty or there is an onto function from $\N$ to~$A$,
(iii)~there is a one-to-one function from $A$ to~$\N$.
\begin{answer}
We will show that (i) implies~(ii), which implies~(iii), and which
......@@ -180,28 +190,32 @@ implies~(i).
First the (i) implies~(ii) proof.
If~$A$ is countable then there are two cases: either~$A$ is finite
or~$A$ is denumerable.
The case that~$A$ is denumerable by definition has a
correspondence between~$\N$ and~$A$.
If~$A$ is finite then it is either empty or nonempty.
The empty case is trivial so suppose that $A$ is finite and nonempty
and let $a$ be an element of~$A$.
Then there is a correspondence~$\map{f}{\set{0,1,\ldots,|A|-1}}{A}$,
The empty set is part of statement~(ii)
so suppose that $A$ is finite and nonempty, and let
$a$ be an element of~$A$.
Then there is a correspondence from some initial segment of the natural
numbers~$\map{f}{\set{0,1,\ldots,|A|-1}}{A}$,
so define $\map{\hat{f}}{\N}{A}$ extending~$f$ by
$\hat{f}(i)=f(i)$ if $i\in\set{0,1,\ldots,|A|-1}$, and
$\hat{f}(i)=a$ otherwise.
The extension~$\hat{f}$ is onto because $f$ is onto.
The other case, that~$A$ is denumerable, by definition has a
correspondence between~$\N$ and~$A$.
$\hat{f}(i)=a$ if $i>|A|-1$.
This extension~$\hat{f}$ is onto because $f$ is onto.
Next is the (ii) implies~(iii) proof.
Next is the proof that statement~(ii) implies~(iii).
By~(ii) either $A$ is the empty set
or there is an onto function from $\N$ to~$A$.
If~$A$ is empty then the empty function is a map with domain~$A$ and
codomain~$\N$ that is vacuously one-to-one.
The other possibility is that there is an onto function~$\map{f}{\N}{A}$.
codomain~$\N$ that is one-to-one, vacuously.
If $A$ is nonempty then there is an onto function~$\map{f}{\N}{A}$.
To get a map~$g$ with domain~$A$ and codomain~$\N$ that is one-to-one
define~$g(a)$ to be the least number $n\in\N$ such that~$f(n)=a$;
at least one such number exists because~$f$ is onto.
Clearly~$g$ is one-to-one.
Finally for the (iii) implies~(i) proof.
Let~$\map{f}{A}{\N}$ be one-to-one.
Finally, the (iii) implies~(i) proof.
Assume that there is a one-to-one function $\map{f}{A}{\N}$.
Let $B=\range(A)\subseteq\N$.
Then the function~$\map{\hat{f}}{A}{B}$ given by~$\hat{f}(a)=f(a)$ is a
correspondence (it only eliminates unused elements of the codomain).
......@@ -215,7 +229,7 @@ Recall that~$B\subseteq\N$ so we can define a map
$\map{g}{\N}{B}$ by:
take $g(j)$ to be the $j$-th largest element of~$B$.
This map is onto because of the assumption that~$B$ is not finite, and it
is one-to-one by construction (if $g(j_0)=g(j_1)$ then the $j_0$-th largest
is one-to-one by construction:~if $g(j_0)=g(j_1)$ then the $j_0$-th largest
element of~$B$ is the same as the $j_1$-th largest element, which is only
possible if $j_0=j_1$).
\end{answer}
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