Mixing
Proving sets of functions have same cardinality
$begingroup$
Prove that $$card(A^{B times C})=card(A^{B^C})$$
where $A^B$ is a set of all functions from $B$ to $A$ and $A times B$ is cartesian product of sets.
Is the bijection supposed to be sending $f$ to $g$, where $g$ is sending $h$ to $t$, or something like that ?
Can't work it out.
functions cardinals
$endgroup$
add a comment |
$begingroup$
Prove that $$card(A^{B times C})=card(A^{B^C})$$
where $A^B$ is a set of all functions from $B$ to $A$ and $A times B$ is cartesian product of sets.
Is the bijection supposed to be sending $f$ to $g$, where $g$ is sending $h$ to $t$, or something like that ?
Can't work it out.
functions cardinals
$endgroup$
add a comment |
$begingroup$
Prove that $$card(A^{B times C})=card(A^{B^C})$$
where $A^B$ is a set of all functions from $B$ to $A$ and $A times B$ is cartesian product of sets.
Is the bijection supposed to be sending $f$ to $g$, where $g$ is sending $h$ to $t$, or something like that ?
Can't work it out.
functions cardinals
$endgroup$
Prove that $$card(A^{B times C})=card(A^{B^C})$$
where $A^B$ is a set of all functions from $B$ to $A$ and $A times B$ is cartesian product of sets.
Is the bijection supposed to be sending $f$ to $g$, where $g$ is sending $h$ to $t$, or something like that ?
Can't work it out.
functions cardinals
functions cardinals
asked Jan 22 at 21:17
user626177
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that with $A^{(B^C)}$, the claim is false - we want ${(A^B)}^C$ (which we recall to indeed equal $A^{BC}$ in the realm of natural numbers).
Given $fcolon Btimes Cto A$, we want to find $gcolon Cto A^B$, i.e., $g$ should, for each $cin C$, produce a map $g(c)colon Bto A$. The choice $g(c)(b)=f(b,c)$ suggests itself.
This correspondence invertible, i.e., given $gin {(A^B)}^C$, we can define $fcolon Btimes Cto A$ by letting $f(b,c)=g(c)(b)$.
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Note that with $A^{(B^C)}$, the claim is false - we want ${(A^B)}^C$ (which we recall to indeed equal $A^{BC}$ in the realm of natural numbers).
Given $fcolon Btimes Cto A$, we want to find $gcolon Cto A^B$, i.e., $g$ should, for each $cin C$, produce a map $g(c)colon Bto A$. The choice $g(c)(b)=f(b,c)$ suggests itself.
This correspondence invertible, i.e., given $gin {(A^B)}^C$, we can define $fcolon Btimes Cto A$ by letting $f(b,c)=g(c)(b)$.
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
add a comment |
$begingroup$
Note that with $A^{(B^C)}$, the claim is false - we want ${(A^B)}^C$ (which we recall to indeed equal $A^{BC}$ in the realm of natural numbers).
Given $fcolon Btimes Cto A$, we want to find $gcolon Cto A^B$, i.e., $g$ should, for each $cin C$, produce a map $g(c)colon Bto A$. The choice $g(c)(b)=f(b,c)$ suggests itself.
This correspondence invertible, i.e., given $gin {(A^B)}^C$, we can define $fcolon Btimes Cto A$ by letting $f(b,c)=g(c)(b)$.
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
add a comment |
$begingroup$
Note that with $A^{(B^C)}$, the claim is false - we want ${(A^B)}^C$ (which we recall to indeed equal $A^{BC}$ in the realm of natural numbers).
Given $fcolon Btimes Cto A$, we want to find $gcolon Cto A^B$, i.e., $g$ should, for each $cin C$, produce a map $g(c)colon Bto A$. The choice $g(c)(b)=f(b,c)$ suggests itself.
This correspondence invertible, i.e., given $gin {(A^B)}^C$, we can define $fcolon Btimes Cto A$ by letting $f(b,c)=g(c)(b)$.
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
Note that with $A^{(B^C)}$, the claim is false - we want ${(A^B)}^C$ (which we recall to indeed equal $A^{BC}$ in the realm of natural numbers).
Given $fcolon Btimes Cto A$, we want to find $gcolon Cto A^B$, i.e., $g$ should, for each $cin C$, produce a map $g(c)colon Bto A$. The choice $g(c)(b)=f(b,c)$ suggests itself.
This correspondence invertible, i.e., given $gin {(A^B)}^C$, we can define $fcolon Btimes Cto A$ by letting $f(b,c)=g(c)(b)$.
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
answered Jan 22 at 21:24
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
add a comment |
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
$begingroup$
Thanks! Altough, had me there confused with notation $g(c)$ for a second.
$endgroup$
– user626177
Jan 22 at 21:34
add a comment |
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