Bijection from 2^X x Y and (2^X)^Y and describe the elements of each
$begingroup$
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
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
$endgroup$
add a comment |
$begingroup$
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
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
$endgroup$
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
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
asked Jan 7 at 13:40
Richard CameronRichard Cameron
121
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For each function $f:Xtimes Yrightarrow {0,1}$ define the
function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
Then $g(y)(x) =f(x,y)$.
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
$endgroup$
add a comment |
$begingroup$
Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 7 at 14:29
drhabdrhab
103k545136
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For each function $f:Xtimes Yrightarrow {0,1}$ define the
function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
Then $g(y)(x) =f(x,y)$.
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
$endgroup$
add a comment |
$begingroup$
For each function $f:Xtimes Yrightarrow {0,1}$ define the
function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
Then $g(y)(x) =f(x,y)$.
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
$endgroup$
add a comment |
$begingroup$
For each function $f:Xtimes Yrightarrow {0,1}$ define the
function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
Then $g(y)(x) =f(x,y)$.
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
$endgroup$
For each function $f:Xtimes Yrightarrow {0,1}$ define the
function $g:Yrightarrow (Xrightarrow{0,1})$ by parametrization $g(y) = f(.,y)$.
Then $g(y)(x) =f(x,y)$.
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
answered Jan 7 at 13:50
WuestenfuxWuestenfux
5,3231513
5,3231513
add a comment |
add a comment |
$begingroup$
Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
Hint: $(y,(x, 1/2 pm 1/2)) mapsto (x, y, 1/2 pm 1/2)$.
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
answered Jan 7 at 13:50
Lucas HenriqueLucas Henrique
1,031414
1,031414
add a comment |
add a comment |
$begingroup$
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).
answered Jan 7 at 14:29
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 7 at 14:29
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 7 at 14:29
drhabdrhab
103k545136
$endgroup$
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).
answered Jan 7 at 14:29
drhabdrhab
103k545136
answered Jan 7 at 14:29
drhabdrhab
103k545136
answered Jan 7 at 14:29
drhabdrhab
103k545136
answered Jan 7 at 14:29
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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$begingroup$
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