Bijection from 2^X x Y and (2^X)^Y and describe the elements of each












-1












$begingroup$


Let 2 = {0,1} and X,Y be sets.



(i) Describe the elements of 2X x Y and (2X)Y



The first is a function from X x Y to 2 so the elements would have the form ((x,y),0) or and ((x,y),1) and the second is a function from Y to the function from X to 2 so the elements would take the form (y,(x,0)) and          .(y,(x,1)).



Is this how I would describe the elements?



(ii) Find a bijective function from 2X x Y and (2X)Y



I know what a bijection is but I'm not sure how to go about finding one and proving it






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    The second case is wrong: elements of $(2^X)^Y$ take the form $(y,f)$, where $fin 2^X$. Your $(x,0)$ and $(x,1)$ are not elements of $2^X$; they are elements of elements of $2^X$.
    $endgroup$
    – TonyK
    Jan 7 at 14:04


















-1












$begingroup$


Let 2 = {0,1} and X,Y be sets.



(i) Describe the elements of 2X x Y and (2X)Y



The first is a function from X x Y to 2 so the elements would have the form ((x,y),0) or and ((x,y),1) and the second is a function from Y to the function from X to 2 so the elements would take the form (y,(x,0)) and          .(y,(x,1)).



Is this how I would describe the elements?



(ii) Find a bijective function from 2X x Y and (2X)Y



I know what a bijection is but I'm not sure how to go about finding one and proving it






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    The second case is wrong: elements of $(2^X)^Y$ take the form $(y,f)$, where $fin 2^X$. Your $(x,0)$ and $(x,1)$ are not elements of $2^X$; they are elements of elements of $2^X$.
    $endgroup$
    – TonyK
    Jan 7 at 14:04
















-1












-1








-1





$begingroup$


Let 2 = {0,1} and X,Y be sets.



(i) Describe the elements of 2X x Y and (2X)Y



The first is a function from X x Y to 2 so the elements would have the form ((x,y),0) or and ((x,y),1) and the second is a function from Y to the function from X to 2 so the elements would take the form (y,(x,0)) and          .(y,(x,1)).



Is this how I would describe the elements?



(ii) Find a bijective function from 2X x Y and (2X)Y



I know what a bijection is but I'm not sure how to go about finding one and proving it






elementary-set-theory







share|cite|improve this question









$endgroup$




Let 2 = {0,1} and X,Y be sets.



(i) Describe the elements of 2X x Y and (2X)Y



The first is a function from X x Y to 2 so the elements would have the form ((x,y),0) or and ((x,y),1) and the second is a function from Y to the function from X to 2 so the elements would take the form (y,(x,0)) and          .(y,(x,1)).



Is this how I would describe the elements?



(ii) Find a bijective function from 2X x Y and (2X)Y



I know what a bijection is but I'm not sure how to go about finding one and proving it






elementary-set-theory




elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 13:40









Richard CameronRichard Cameron

121




121












  • $begingroup$
    The second case is wrong: elements of $(2^X)^Y$ take the form $(y,f)$, where $fin 2^X$. Your $(x,0)$ and $(x,1)$ are not elements of $2^X$; they are elements of elements of $2^X$.
    $endgroup$
    – TonyK
    Jan 7 at 14:04




















  • $begingroup$
    The second case is wrong: elements of $(2^X)^Y$ take the form $(y,f)$, where $fin 2^X$. Your $(x,0)$ and $(x,1)$ are not elements of $2^X$; they are elements of elements of $2^X$.
    $endgroup$
    – TonyK
    Jan 7 at 14:04


















$begingroup$
The second case is wrong: elements of $(2^X)^Y$ take the form $(y,f)$, where $fin 2^X$. Your $(x,0)$ and $(x,1)$ are not elements of $2^X$; they are elements of elements of $2^X$.
$endgroup$
– TonyK
Jan 7 at 14:04






$begingroup$
The second case is wrong: elements of $(2^X)^Y$ take the form $(y,f)$, where $fin 2^X$. Your $(x,0)$ and $(x,1)$ are not elements of $2^X$; they are elements of elements of $2^X$.
$endgroup$
– TonyK
Jan 7 at 14:04












3 Answers
3






active

oldest

votes


















1












$begingroup$

For each function $f:Xtimes Yrightarrow {0,1}$ define the
function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
Then $g(y)(x) =f(x,y)$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Alternative approach.



      Identify $2^{Xtimes Y}$ with $wp(Xtimes Y)$ and identify $left(2^Xright)^Y$ with ${fmid ftext{ is a function }Ytowp(X)}$.




      There is a one-to-one relation between subsets of $Xtimes Y$ and functions $Ytowp(X)$.



      If $Asubseteq Xtimes Y$ then $A$ determines the function $f_A:Ytowp(X)$ that is prescribed by:$$ymapsto{xin Xmidlangle x,yranglein A}$$



      If conversely $f:Ytowp(X)$ denotes a function then $f$ determines the set:$$A_f:={langle x,yranglemid yin Xwedge xin f(y)}$$



      It is not difficult to prove that $f_{A_f}=f$ and $A_{f_A}=A$ (proving that the relation is one to one).






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        For each function $f:Xtimes Yrightarrow {0,1}$ define the
        function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
        Then $g(y)(x) =f(x,y)$.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          For each function $f:Xtimes Yrightarrow {0,1}$ define the
          function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
          Then $g(y)(x) =f(x,y)$.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            For each function $f:Xtimes Yrightarrow {0,1}$ define the
            function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
            Then $g(y)(x) =f(x,y)$.






            share|cite|improve this answer









            $endgroup$



            For each function $f:Xtimes Yrightarrow {0,1}$ define the
            function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
            Then $g(y)(x) =f(x,y)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 13:50









            WuestenfuxWuestenfux

            5,3231513




            5,3231513























                0












                $begingroup$

                Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 13:50









                    Lucas HenriqueLucas Henrique

                    1,031414




                    1,031414























                        0












                        $begingroup$

                        Alternative approach.



                        Identify $2^{Xtimes Y}$ with $wp(Xtimes Y)$ and identify $left(2^Xright)^Y$ with ${fmid ftext{ is a function }Ytowp(X)}$.




                        There is a one-to-one relation between subsets of $Xtimes Y$ and functions $Ytowp(X)$.



                        If $Asubseteq Xtimes Y$ then $A$ determines the function $f_A:Ytowp(X)$ that is prescribed by:$$ymapsto{xin Xmidlangle x,yranglein A}$$



                        If conversely $f:Ytowp(X)$ denotes a function then $f$ determines the set:$$A_f:={langle x,yranglemid yin Xwedge xin f(y)}$$



                        It is not difficult to prove that $f_{A_f}=f$ and $A_{f_A}=A$ (proving that the relation is one to one).






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Alternative approach.



                          Identify $2^{Xtimes Y}$ with $wp(Xtimes Y)$ and identify $left(2^Xright)^Y$ with ${fmid ftext{ is a function }Ytowp(X)}$.




                          There is a one-to-one relation between subsets of $Xtimes Y$ and functions $Ytowp(X)$.



                          If $Asubseteq Xtimes Y$ then $A$ determines the function $f_A:Ytowp(X)$ that is prescribed by:$$ymapsto{xin Xmidlangle x,yranglein A}$$



                          If conversely $f:Ytowp(X)$ denotes a function then $f$ determines the set:$$A_f:={langle x,yranglemid yin Xwedge xin f(y)}$$



                          It is not difficult to prove that $f_{A_f}=f$ and $A_{f_A}=A$ (proving that the relation is one to one).






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Alternative approach.



                            Identify $2^{Xtimes Y}$ with $wp(Xtimes Y)$ and identify $left(2^Xright)^Y$ with ${fmid ftext{ is a function }Ytowp(X)}$.




                            There is a one-to-one relation between subsets of $Xtimes Y$ and functions $Ytowp(X)$.



                            If $Asubseteq Xtimes Y$ then $A$ determines the function $f_A:Ytowp(X)$ that is prescribed by:$$ymapsto{xin Xmidlangle x,yranglein A}$$



                            If conversely $f:Ytowp(X)$ denotes a function then $f$ determines the set:$$A_f:={langle x,yranglemid yin Xwedge xin f(y)}$$



                            It is not difficult to prove that $f_{A_f}=f$ and $A_{f_A}=A$ (proving that the relation is one to one).






                            share|cite|improve this answer









                            $endgroup$



                            Alternative approach.



                            Identify $2^{Xtimes Y}$ with $wp(Xtimes Y)$ and identify $left(2^Xright)^Y$ with ${fmid ftext{ is a function }Ytowp(X)}$.




                            There is a one-to-one relation between subsets of $Xtimes Y$ and functions $Ytowp(X)$.



                            If $Asubseteq Xtimes Y$ then $A$ determines the function $f_A:Ytowp(X)$ that is prescribed by:$$ymapsto{xin Xmidlangle x,yranglein A}$$



                            If conversely $f:Ytowp(X)$ denotes a function then $f$ determines the set:$$A_f:={langle x,yranglemid yin Xwedge xin f(y)}$$



                            It is not difficult to prove that $f_{A_f}=f$ and $A_{f_A}=A$ (proving that the relation is one to one).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 7 at 14:29









                            drhabdrhab

                            103k545136




                            103k545136






























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