Mixing
Defining a tricky function: $(A rightarrow mathcal{P}(B)) rightarrowmathcal{P}(A rightarrow B)$
$begingroup$
How would I define a function of the form:
begin{align*}
phi: (A rightarrow mathcal{P}(B)) rightarrowmathcal{P}(A rightarrow B)
end{align*}
I know what behaviour I want, I'm just struggling to define it. For example,
consider a function
begin{align*}
&f: {0, 1} rightarrow mathcal{P}({0, 1, 2, 3}) \
&f(0) = {0, 1} \
&f(1) = {2, 3} \
end{align*}
I want $phi(f)$ to be:
begin{align*}
&phi(f) = {g_1, g_2, g_3, g_4 } \
&g_1(0) = 0 qquad g_1(1) = 2 \
&g_2(0) = 0 qquad g_2(1) = 3 \
&g_3(0) = 1 qquad g_3(1) = 2 \
&g_4(0) = 1 qquad g_4(2) = 3 \
end{align*}
That is, I want all possible combinations of ${0, 1} times {2, 3}$ as functions.
functions elementary-set-theory
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
$endgroup$
add a comment |
$begingroup$
How would I define a function of the form:
begin{align*}
phi: (A rightarrow mathcal{P}(B)) rightarrowmathcal{P}(A rightarrow B)
end{align*}
I know what behaviour I want, I'm just struggling to define it. For example,
consider a function
begin{align*}
&f: {0, 1} rightarrow mathcal{P}({0, 1, 2, 3}) \
&f(0) = {0, 1} \
&f(1) = {2, 3} \
end{align*}
I want $phi(f)$ to be:
begin{align*}
&phi(f) = {g_1, g_2, g_3, g_4 } \
&g_1(0) = 0 qquad g_1(1) = 2 \
&g_2(0) = 0 qquad g_2(1) = 3 \
&g_3(0) = 1 qquad g_3(1) = 2 \
&g_4(0) = 1 qquad g_4(2) = 3 \
end{align*}
That is, I want all possible combinations of ${0, 1} times {2, 3}$ as functions.
functions elementary-set-theory
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
$endgroup$
2
$begingroup$
It seems to me that you are asking about Cartesian product: if $(X_a)_{ain A}$ is a family of sets indexed by $A$, then $$ prod_{a in A} X_a = left{ g : A to bigcup_{a in A} X_a , middle| , g(a) in X_a text{ for all } a in A right}. $$ In such case, $phi(f) = prod_{a in A} f(a)$.
$endgroup$
– Sangchul Lee
Jan 24 at 21:52
$begingroup$
Is there anyway you could rewrite this so that you aren't reusing the symbols $0$ and $1$? It just makes trying to figure out what you are asking for harder.
$endgroup$
– DanielV
Jan 24 at 22:16
add a comment |
$begingroup$
How would I define a function of the form:
begin{align*}
phi: (A rightarrow mathcal{P}(B)) rightarrowmathcal{P}(A rightarrow B)
end{align*}
I know what behaviour I want, I'm just struggling to define it. For example,
consider a function
begin{align*}
&f: {0, 1} rightarrow mathcal{P}({0, 1, 2, 3}) \
&f(0) = {0, 1} \
&f(1) = {2, 3} \
end{align*}
I want $phi(f)$ to be:
begin{align*}
&phi(f) = {g_1, g_2, g_3, g_4 } \
&g_1(0) = 0 qquad g_1(1) = 2 \
&g_2(0) = 0 qquad g_2(1) = 3 \
&g_3(0) = 1 qquad g_3(1) = 2 \
&g_4(0) = 1 qquad g_4(2) = 3 \
end{align*}
That is, I want all possible combinations of ${0, 1} times {2, 3}$ as functions.
functions elementary-set-theory
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
$endgroup$
How would I define a function of the form:
begin{align*}
phi: (A rightarrow mathcal{P}(B)) rightarrowmathcal{P}(A rightarrow B)
end{align*}
I know what behaviour I want, I'm just struggling to define it. For example,
consider a function
begin{align*}
&f: {0, 1} rightarrow mathcal{P}({0, 1, 2, 3}) \
&f(0) = {0, 1} \
&f(1) = {2, 3} \
end{align*}
I want $phi(f)$ to be:
begin{align*}
&phi(f) = {g_1, g_2, g_3, g_4 } \
&g_1(0) = 0 qquad g_1(1) = 2 \
&g_2(0) = 0 qquad g_2(1) = 3 \
&g_3(0) = 1 qquad g_3(1) = 2 \
&g_4(0) = 1 qquad g_4(2) = 3 \
end{align*}
That is, I want all possible combinations of ${0, 1} times {2, 3}$ as functions.
functions elementary-set-theory
functions elementary-set-theory
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
asked Jan 24 at 21:30
Siddharth BhatSiddharth Bhat
2,9121918
2,9121918
2
$begingroup$
It seems to me that you are asking about Cartesian product: if $(X_a)_{ain A}$ is a family of sets indexed by $A$, then $$ prod_{a in A} X_a = left{ g : A to bigcup_{a in A} X_a , middle| , g(a) in X_a text{ for all } a in A right}. $$ In such case, $phi(f) = prod_{a in A} f(a)$.
$endgroup$
– Sangchul Lee
Jan 24 at 21:52
$begingroup$
Is there anyway you could rewrite this so that you aren't reusing the symbols $0$ and $1$? It just makes trying to figure out what you are asking for harder.
$endgroup$
– DanielV
Jan 24 at 22:16
add a comment |
2
$begingroup$
It seems to me that you are asking about Cartesian product: if $(X_a)_{ain A}$ is a family of sets indexed by $A$, then $$ prod_{a in A} X_a = left{ g : A to bigcup_{a in A} X_a , middle| , g(a) in X_a text{ for all } a in A right}. $$ In such case, $phi(f) = prod_{a in A} f(a)$.
$endgroup$
– Sangchul Lee
Jan 24 at 21:52
$begingroup$
Is there anyway you could rewrite this so that you aren't reusing the symbols $0$ and $1$? It just makes trying to figure out what you are asking for harder.
$endgroup$
– DanielV
Jan 24 at 22:16
2
2
$begingroup$
It seems to me that you are asking about Cartesian product: if $(X_a)_{ain A}$ is a family of sets indexed by $A$, then $$ prod_{a in A} X_a = left{ g : A to bigcup_{a in A} X_a , middle| , g(a) in X_a text{ for all } a in A right}. $$ In such case, $phi(f) = prod_{a in A} f(a)$.
$endgroup$
– Sangchul Lee
Jan 24 at 21:52
$begingroup$
It seems to me that you are asking about Cartesian product: if $(X_a)_{ain A}$ is a family of sets indexed by $A$, then $$ prod_{a in A} X_a = left{ g : A to bigcup_{a in A} X_a , middle| , g(a) in X_a text{ for all } a in A right}. $$ In such case, $phi(f) = prod_{a in A} f(a)$.
$endgroup$
– Sangchul Lee
Jan 24 at 21:52
$begingroup$
Is there anyway you could rewrite this so that you aren't reusing the symbols $0$ and $1$? It just makes trying to figure out what you are asking for harder.
$endgroup$
– DanielV
Jan 24 at 22:16
$begingroup$
Is there anyway you could rewrite this so that you aren't reusing the symbols $0$ and $1$? It just makes trying to figure out what you are asking for harder.
$endgroup$
– DanielV
Jan 24 at 22:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The elements of $phi(f)$ are exactly those $g colon A to B$ for which
$$forall a in A: g(a) in f(a).$$
They are choice functions: for every $a in A$, they choose an element $g(a) in f(a)$. The fact such a function, in general, exists at all is the axiom of choice.
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
$endgroup$
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
1
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
add a comment |
$begingroup$
This is not properly speaking an answer to the question ; it is an attempt to see more clearly what all this is about.
I will use the term "application" as a synonym to a function $f:E to F$ such that every $e in E$ has an image $f(e) in F$.
You must know the following notations :
$F^{E}$ for the set of applications from $E$ to $F$ and, as a particular case
${0,1}^G cong 2^G$ for the (well named) power set, i.e., the set of subsets of set $G$.
(see for example : https://math.stackexchange.com/q/104524)
If I understand well, you are asking whether there is a (more or less natural) way to find a "path" between :
$ S_1:=(2^B)^{A}cong 2^{B times A}$ and $S_2=2^{(B^{A})}.$
The first isomorphism drives us to the partitions of the cartesian product $B times A$ or the isomorphic set $A times B$. Among such partitions, there are those which define an application from $A$ to $B$, i.e., a partition that is ${(a_1,b_1),(a_2,b_2),...(a_n,b_n)}$ where ${a_1,a_2,...a_n}=A$ (every element of set $A$ is present exactly once). Thus, in a certain sense, the set of applications $f:A to B$ is included into $S_1$.
Concerning $S_2=2^{(B^{A})}$, it could be described as the ways to group applications. It is a much "larger" set than $S_1$, and I don't see any clearcut correspondence between $S_1$ and $S_2$.
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The elements of $phi(f)$ are exactly those $g colon A to B$ for which
$$forall a in A: g(a) in f(a).$$
They are choice functions: for every $a in A$, they choose an element $g(a) in f(a)$. The fact such a function, in general, exists at all is the axiom of choice.
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
$endgroup$
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
1
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
add a comment |
$begingroup$
The elements of $phi(f)$ are exactly those $g colon A to B$ for which
$$forall a in A: g(a) in f(a).$$
They are choice functions: for every $a in A$, they choose an element $g(a) in f(a)$. The fact such a function, in general, exists at all is the axiom of choice.
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
$endgroup$
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
1
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
add a comment |
$begingroup$
The elements of $phi(f)$ are exactly those $g colon A to B$ for which
$$forall a in A: g(a) in f(a).$$
They are choice functions: for every $a in A$, they choose an element $g(a) in f(a)$. The fact such a function, in general, exists at all is the axiom of choice.
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
$endgroup$
The elements of $phi(f)$ are exactly those $g colon A to B$ for which
$$forall a in A: g(a) in f(a).$$
They are choice functions: for every $a in A$, they choose an element $g(a) in f(a)$. The fact such a function, in general, exists at all is the axiom of choice.
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
edited Jan 25 at 20:56
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
answered Jan 24 at 22:09
MagdiragdagMagdiragdag
11.1k31532
11.1k31532
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
1
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
add a comment |
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
1
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
$begingroup$
The generality of invoking choice is only necessary for infinite products, which is not the case here. Is there some slick way to define the family of choice functions?
$endgroup$
– Siddharth Bhat
Jan 25 at 19:45
1
1
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
$begingroup$
It's not clear from your question that you want $A$ and $B$ to be finite. Anyway, that doesn't matter for the definition of $phi(f)$. It still is ${ g colon A to B ;|; forall a in A: g(a) in f(a)}$. For finite $A$ and $B$, that set contains $prod_{a in A} |f(a)|$ elements. You only need choice to prove that (in the general case, not restricting to finite $A$ and $B$), the set is non-empty (assuming $f(a)$ is never empty, of course).
$endgroup$
– Magdiragdag
Jan 25 at 20:46
add a comment |
$begingroup$
This is not properly speaking an answer to the question ; it is an attempt to see more clearly what all this is about.
I will use the term "application" as a synonym to a function $f:E to F$ such that every $e in E$ has an image $f(e) in F$.
You must know the following notations :
$F^{E}$ for the set of applications from $E$ to $F$ and, as a particular case
${0,1}^G cong 2^G$ for the (well named) power set, i.e., the set of subsets of set $G$.
(see for example : https://math.stackexchange.com/q/104524)
If I understand well, you are asking whether there is a (more or less natural) way to find a "path" between :
$ S_1:=(2^B)^{A}cong 2^{B times A}$ and $S_2=2^{(B^{A})}.$
The first isomorphism drives us to the partitions of the cartesian product $B times A$ or the isomorphic set $A times B$. Among such partitions, there are those which define an application from $A$ to $B$, i.e., a partition that is ${(a_1,b_1),(a_2,b_2),...(a_n,b_n)}$ where ${a_1,a_2,...a_n}=A$ (every element of set $A$ is present exactly once). Thus, in a certain sense, the set of applications $f:A to B$ is included into $S_1$.
Concerning $S_2=2^{(B^{A})}$, it could be described as the ways to group applications. It is a much "larger" set than $S_1$, and I don't see any clearcut correspondence between $S_1$ and $S_2$.
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
$endgroup$
add a comment |
$begingroup$
This is not properly speaking an answer to the question ; it is an attempt to see more clearly what all this is about.
I will use the term "application" as a synonym to a function $f:E to F$ such that every $e in E$ has an image $f(e) in F$.
You must know the following notations :
$F^{E}$ for the set of applications from $E$ to $F$ and, as a particular case
${0,1}^G cong 2^G$ for the (well named) power set, i.e., the set of subsets of set $G$.
(see for example : https://math.stackexchange.com/q/104524)
If I understand well, you are asking whether there is a (more or less natural) way to find a "path" between :
$ S_1:=(2^B)^{A}cong 2^{B times A}$ and $S_2=2^{(B^{A})}.$
The first isomorphism drives us to the partitions of the cartesian product $B times A$ or the isomorphic set $A times B$. Among such partitions, there are those which define an application from $A$ to $B$, i.e., a partition that is ${(a_1,b_1),(a_2,b_2),...(a_n,b_n)}$ where ${a_1,a_2,...a_n}=A$ (every element of set $A$ is present exactly once). Thus, in a certain sense, the set of applications $f:A to B$ is included into $S_1$.
Concerning $S_2=2^{(B^{A})}$, it could be described as the ways to group applications. It is a much "larger" set than $S_1$, and I don't see any clearcut correspondence between $S_1$ and $S_2$.
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
$endgroup$
add a comment |
$begingroup$
This is not properly speaking an answer to the question ; it is an attempt to see more clearly what all this is about.
I will use the term "application" as a synonym to a function $f:E to F$ such that every $e in E$ has an image $f(e) in F$.
You must know the following notations :
$F^{E}$ for the set of applications from $E$ to $F$ and, as a particular case
${0,1}^G cong 2^G$ for the (well named) power set, i.e., the set of subsets of set $G$.
(see for example : https://math.stackexchange.com/q/104524)
If I understand well, you are asking whether there is a (more or less natural) way to find a "path" between :
$ S_1:=(2^B)^{A}cong 2^{B times A}$ and $S_2=2^{(B^{A})}.$
The first isomorphism drives us to the partitions of the cartesian product $B times A$ or the isomorphic set $A times B$. Among such partitions, there are those which define an application from $A$ to $B$, i.e., a partition that is ${(a_1,b_1),(a_2,b_2),...(a_n,b_n)}$ where ${a_1,a_2,...a_n}=A$ (every element of set $A$ is present exactly once). Thus, in a certain sense, the set of applications $f:A to B$ is included into $S_1$.
Concerning $S_2=2^{(B^{A})}$, it could be described as the ways to group applications. It is a much "larger" set than $S_1$, and I don't see any clearcut correspondence between $S_1$ and $S_2$.
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
$endgroup$
This is not properly speaking an answer to the question ; it is an attempt to see more clearly what all this is about.
I will use the term "application" as a synonym to a function $f:E to F$ such that every $e in E$ has an image $f(e) in F$.
You must know the following notations :
$F^{E}$ for the set of applications from $E$ to $F$ and, as a particular case
${0,1}^G cong 2^G$ for the (well named) power set, i.e., the set of subsets of set $G$.
(see for example : https://math.stackexchange.com/q/104524)
If I understand well, you are asking whether there is a (more or less natural) way to find a "path" between :
$ S_1:=(2^B)^{A}cong 2^{B times A}$ and $S_2=2^{(B^{A})}.$
The first isomorphism drives us to the partitions of the cartesian product $B times A$ or the isomorphic set $A times B$. Among such partitions, there are those which define an application from $A$ to $B$, i.e., a partition that is ${(a_1,b_1),(a_2,b_2),...(a_n,b_n)}$ where ${a_1,a_2,...a_n}=A$ (every element of set $A$ is present exactly once). Thus, in a certain sense, the set of applications $f:A to B$ is included into $S_1$.
Concerning $S_2=2^{(B^{A})}$, it could be described as the ways to group applications. It is a much "larger" set than $S_1$, and I don't see any clearcut correspondence between $S_1$ and $S_2$.
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
edited Jan 24 at 22:44
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
answered Jan 24 at 22:25
Jean MarieJean Marie
30.7k42154
30.7k42154
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2
$begingroup$
It seems to me that you are asking about Cartesian product: if $(X_a)_{ain A}$ is a family of sets indexed by $A$, then $$ prod_{a in A} X_a = left{ g : A to bigcup_{a in A} X_a , middle| , g(a) in X_a text{ for all } a in A right}. $$ In such case, $phi(f) = prod_{a in A} f(a)$.
$endgroup$
– Sangchul Lee
Jan 24 at 21:52
$begingroup$
Is there anyway you could rewrite this so that you aren't reusing the symbols $0$ and $1$? It just makes trying to figure out what you are asking for harder.
$endgroup$
– DanielV
Jan 24 at 22:16