Prove that (if $R$ is set-like, then its transitive closure is set-like) implies the axiom of replacement
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This is exercise I.9.6 from Kunen's Set Theory (2011). $R$ is set-like on the class $A$ iff $y in A$ implies ${ x in A: xRy }$ is a set. I am not so sure about this, but it is implied that this can be done without using the axiom of choice.
The obvious choice of $R$ is $xRy$ iff $x= F(y)$, however what I am most unsure of is how to pick the class such that it cointains all of $A$ and all of $F(A)$. I think the simple $A cup F(A)$ doesn't work because for an $x in A$, $F(F(x))$ may not be a set and hence $R$ wouldn't be set-like. This reasoning seems to apply to any set $B$, since $A cup F(A)$ must be cointained in $B$ in order for the transitive closure to include $F(A)$.
As an specific example, consider the set $mathbb{N}$ of natural numbers and the formula $phi(x,y)$ "$y$ is $mathbb{N}$ if $x$ is a natural number and, $y$ is ${w: w=w}$ if $x$ is $mathbb{N}$". Clearly while $R$ is set-like in $mathbb{N}$, it is not set-like in $mathbb{N} cup F(mathbb{N})$, and in general I can't modify $phi(x,y)$ as to exclude the $x$s in $F(A)$, as they may be part of $A$, too.
So, how can we prove it?
set-theory
asked Jan 8 at 20:36
RyunaqRyunaq
957
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add a comment |
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This is exercise I.9.6 from Kunen's Set Theory (2011). $R$ is set-like on the class $A$ iff $y in A$ implies ${ x in A: xRy }$ is a set. I am not so sure about this, but it is implied that this can be done without using the axiom of choice.
The obvious choice of $R$ is $xRy$ iff $x= F(y)$, however what I am most unsure of is how to pick the class such that it cointains all of $A$ and all of $F(A)$. I think the simple $A cup F(A)$ doesn't work because for an $x in A$, $F(F(x))$ may not be a set and hence $R$ wouldn't be set-like. This reasoning seems to apply to any set $B$, since $A cup F(A)$ must be cointained in $B$ in order for the transitive closure to include $F(A)$.
As an specific example, consider the set $mathbb{N}$ of natural numbers and the formula $phi(x,y)$ "$y$ is $mathbb{N}$ if $x$ is a natural number and, $y$ is ${w: w=w}$ if $x$ is $mathbb{N}$". Clearly while $R$ is set-like in $mathbb{N}$, it is not set-like in $mathbb{N} cup F(mathbb{N})$, and in general I can't modify $phi(x,y)$ as to exclude the $x$s in $F(A)$, as they may be part of $A$, too.
So, how can we prove it?
set-theory
asked Jan 8 at 20:36
RyunaqRyunaq
957
$endgroup$
add a comment |
$begingroup$
This is exercise I.9.6 from Kunen's Set Theory (2011). $R$ is set-like on the class $A$ iff $y in A$ implies ${ x in A: xRy }$ is a set. I am not so sure about this, but it is implied that this can be done without using the axiom of choice.
The obvious choice of $R$ is $xRy$ iff $x= F(y)$, however what I am most unsure of is how to pick the class such that it cointains all of $A$ and all of $F(A)$. I think the simple $A cup F(A)$ doesn't work because for an $x in A$, $F(F(x))$ may not be a set and hence $R$ wouldn't be set-like. This reasoning seems to apply to any set $B$, since $A cup F(A)$ must be cointained in $B$ in order for the transitive closure to include $F(A)$.
As an specific example, consider the set $mathbb{N}$ of natural numbers and the formula $phi(x,y)$ "$y$ is $mathbb{N}$ if $x$ is a natural number and, $y$ is ${w: w=w}$ if $x$ is $mathbb{N}$". Clearly while $R$ is set-like in $mathbb{N}$, it is not set-like in $mathbb{N} cup F(mathbb{N})$, and in general I can't modify $phi(x,y)$ as to exclude the $x$s in $F(A)$, as they may be part of $A$, too.
So, how can we prove it?
set-theory
asked Jan 8 at 20:36
RyunaqRyunaq
957
$endgroup$
This is exercise I.9.6 from Kunen's Set Theory (2011). $R$ is set-like on the class $A$ iff $y in A$ implies ${ x in A: xRy }$ is a set. I am not so sure about this, but it is implied that this can be done without using the axiom of choice.
The obvious choice of $R$ is $xRy$ iff $x= F(y)$, however what I am most unsure of is how to pick the class such that it cointains all of $A$ and all of $F(A)$. I think the simple $A cup F(A)$ doesn't work because for an $x in A$, $F(F(x))$ may not be a set and hence $R$ wouldn't be set-like. This reasoning seems to apply to any set $B$, since $A cup F(A)$ must be cointained in $B$ in order for the transitive closure to include $F(A)$.
As an specific example, consider the set $mathbb{N}$ of natural numbers and the formula $phi(x,y)$ "$y$ is $mathbb{N}$ if $x$ is a natural number and, $y$ is ${w: w=w}$ if $x$ is $mathbb{N}$". Clearly while $R$ is set-like in $mathbb{N}$, it is not set-like in $mathbb{N} cup F(mathbb{N})$, and in general I can't modify $phi(x,y)$ as to exclude the $x$s in $F(A)$, as they may be part of $A$, too.
So, how can we prove it?
set-theory
set-theory
asked Jan 8 at 20:36
RyunaqRyunaq
957
asked Jan 8 at 20:36
RyunaqRyunaq
957
asked Jan 8 at 20:36
RyunaqRyunaq
957
asked Jan 8 at 20:36
RyunaqRyunaq
957
asked Jan 8 at 20:36
RyunaqRyunaq
957
957
add a comment |
add a comment |
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