...

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} ... ...
No preview for this file type
No preview for this file type