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Commits (1)
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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}} \put(1,0){\chanpoint{1,0}} ... ... @@ -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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