Prove that sets being equipotent is an equivalence relation
$begingroup$
Let $X$ and $Y$ two subsets of $E$. We know that $X$ and $Y$ are equipotent, $Xsim Y$, if there exists a bijective function so that $f:Xrightarrow Y$. Proof that $sim$ defines an equivalence relation in $mathcal{P}(E)$.
I don't know how to do that, please help me!
elementary-set-theory
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ two subsets of $E$. We know that $X$ and $Y$ are equipotent, $Xsim Y$, if there exists a bijective function so that $f:Xrightarrow Y$. Proof that $sim$ defines an equivalence relation in $mathcal{P}(E)$.
I don't know how to do that, please help me!
elementary-set-theory
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
$endgroup$
1
$begingroup$
Have you tried showing if the defining properties of an equivalence relation hold for the given relation?
$endgroup$
– OnceUponACrinoid
Jun 7 '15 at 6:22
$begingroup$
That's the problem, I don't know how to define the relation
$endgroup$
– Fhmos31
Jun 7 '15 at 6:32
add a comment |
$begingroup$
Let $X$ and $Y$ two subsets of $E$. We know that $X$ and $Y$ are equipotent, $Xsim Y$, if there exists a bijective function so that $f:Xrightarrow Y$. Proof that $sim$ defines an equivalence relation in $mathcal{P}(E)$.
I don't know how to do that, please help me!
elementary-set-theory
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
$endgroup$
Let $X$ and $Y$ two subsets of $E$. We know that $X$ and $Y$ are equipotent, $Xsim Y$, if there exists a bijective function so that $f:Xrightarrow Y$. Proof that $sim$ defines an equivalence relation in $mathcal{P}(E)$.
I don't know how to do that, please help me!
elementary-set-theory
elementary-set-theory
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
edited Jun 7 '15 at 8:43
Henno Brandsma
110k347116
110k347116
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
asked Jun 7 '15 at 6:18
Fhmos31Fhmos31
12
12
1
$begingroup$
Have you tried showing if the defining properties of an equivalence relation hold for the given relation?
$endgroup$
– OnceUponACrinoid
Jun 7 '15 at 6:22
$begingroup$
That's the problem, I don't know how to define the relation
$endgroup$
– Fhmos31
Jun 7 '15 at 6:32
add a comment |
1
$begingroup$
Have you tried showing if the defining properties of an equivalence relation hold for the given relation?
$endgroup$
– OnceUponACrinoid
Jun 7 '15 at 6:22
$begingroup$
That's the problem, I don't know how to define the relation
$endgroup$
– Fhmos31
Jun 7 '15 at 6:32
1
1
$begingroup$
Have you tried showing if the defining properties of an equivalence relation hold for the given relation?
$endgroup$
– OnceUponACrinoid
Jun 7 '15 at 6:22
$begingroup$
Have you tried showing if the defining properties of an equivalence relation hold for the given relation?
$endgroup$
– OnceUponACrinoid
Jun 7 '15 at 6:22
$begingroup$
That's the problem, I don't know how to define the relation
$endgroup$
– Fhmos31
Jun 7 '15 at 6:32
$begingroup$
That's the problem, I don't know how to define the relation
$endgroup$
– Fhmos31
Jun 7 '15 at 6:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To see reflexivity, observe that the identity map on $X$ is a bijection. To see symmetry, observe that bijections have inverses, so if $Xsim Y$, then there is a bijection $f:Xto Y$, so there is a bijection $f^{-1}:Yto X$. To see transitivity, observe that the composition of two bijections is a bijection, so if $Xsim Y$ and $Ysim Z$, then there are bijections $f:Xto Y$ and $g:Yto Z$, so you have a bijection $gcirc f:Xto Z$ which implies $Xsim Z$.
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
$endgroup$
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
add a comment |
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$begingroup$
To see reflexivity, observe that the identity map on $X$ is a bijection. To see symmetry, observe that bijections have inverses, so if $Xsim Y$, then there is a bijection $f:Xto Y$, so there is a bijection $f^{-1}:Yto X$. To see transitivity, observe that the composition of two bijections is a bijection, so if $Xsim Y$ and $Ysim Z$, then there are bijections $f:Xto Y$ and $g:Yto Z$, so you have a bijection $gcirc f:Xto Z$ which implies $Xsim Z$.
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
$endgroup$
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
add a comment |
$begingroup$
To see reflexivity, observe that the identity map on $X$ is a bijection. To see symmetry, observe that bijections have inverses, so if $Xsim Y$, then there is a bijection $f:Xto Y$, so there is a bijection $f^{-1}:Yto X$. To see transitivity, observe that the composition of two bijections is a bijection, so if $Xsim Y$ and $Ysim Z$, then there are bijections $f:Xto Y$ and $g:Yto Z$, so you have a bijection $gcirc f:Xto Z$ which implies $Xsim Z$.
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
$endgroup$
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
add a comment |
$begingroup$
To see reflexivity, observe that the identity map on $X$ is a bijection. To see symmetry, observe that bijections have inverses, so if $Xsim Y$, then there is a bijection $f:Xto Y$, so there is a bijection $f^{-1}:Yto X$. To see transitivity, observe that the composition of two bijections is a bijection, so if $Xsim Y$ and $Ysim Z$, then there are bijections $f:Xto Y$ and $g:Yto Z$, so you have a bijection $gcirc f:Xto Z$ which implies $Xsim Z$.
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
$endgroup$
To see reflexivity, observe that the identity map on $X$ is a bijection. To see symmetry, observe that bijections have inverses, so if $Xsim Y$, then there is a bijection $f:Xto Y$, so there is a bijection $f^{-1}:Yto X$. To see transitivity, observe that the composition of two bijections is a bijection, so if $Xsim Y$ and $Ysim Z$, then there are bijections $f:Xto Y$ and $g:Yto Z$, so you have a bijection $gcirc f:Xto Z$ which implies $Xsim Z$.
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
answered Jun 7 '15 at 6:42
FunktoralityFunktorality
1,7531419
1,7531419
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
add a comment |
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
$begingroup$
Great, thank you Stan!
$endgroup$
– Fhmos31
Jun 7 '15 at 6:53
add a comment |
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1
$begingroup$
Have you tried showing if the defining properties of an equivalence relation hold for the given relation?
$endgroup$
– OnceUponACrinoid
Jun 7 '15 at 6:22
$begingroup$
That's the problem, I don't know how to define the relation
$endgroup$
– Fhmos31
Jun 7 '15 at 6:32