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Constructibility of uniquely defined sets in ZFC.
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I was thinking about constructible universe, and I had the following idea.
Suppose we have a predicate $phi$ defined in a language of set theory. Suppose moreover, that the statement "there exists $X$ such that $phi(X)$; and for any $X, Y$ if $phi(X)$ and $phi(Y)$, then $X=Y$" is provable in ZFC (essentially, that the formula $phi$ determines a unique set). Then that set $X$ must belong to some level $L_alpha$ of the constructible hierarchy.
The argument is as follows: since the existence and uniqueness is provable in ZFC, it must be true in the constructible universe, which is a model of ZFC. But since constructible universe is an inner model of ZFC, that particular element of the constructible universe that serves as a witness of the truth of the formula in the constructible universe model of ZFC must also be the unique $X$ in the "whole" ZFC$-$we cannot have a different one, because there is already a set that satisfies $X$ in the constructible hierarchy.
Is this argument sound? It feels solid and completely obvious to me, and vacuous at the same time.
logic set-theory first-order-logic model-theory
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
asked Aug 10 at 21:50
xyzzyz
5,4611220
add a comment |Â
up vote
4
down vote
favorite
I was thinking about constructible universe, and I had the following idea.
Suppose we have a predicate $phi$ defined in a language of set theory. Suppose moreover, that the statement "there exists $X$ such that $phi(X)$; and for any $X, Y$ if $phi(X)$ and $phi(Y)$, then $X=Y$" is provable in ZFC (essentially, that the formula $phi$ determines a unique set). Then that set $X$ must belong to some level $L_alpha$ of the constructible hierarchy.
The argument is as follows: since the existence and uniqueness is provable in ZFC, it must be true in the constructible universe, which is a model of ZFC. But since constructible universe is an inner model of ZFC, that particular element of the constructible universe that serves as a witness of the truth of the formula in the constructible universe model of ZFC must also be the unique $X$ in the "whole" ZFC$-$we cannot have a different one, because there is already a set that satisfies $X$ in the constructible hierarchy.
Is this argument sound? It feels solid and completely obvious to me, and vacuous at the same time.
logic set-theory first-order-logic model-theory
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
asked Aug 10 at 21:50
xyzzyz
5,4611220
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I was thinking about constructible universe, and I had the following idea.
Suppose we have a predicate $phi$ defined in a language of set theory. Suppose moreover, that the statement "there exists $X$ such that $phi(X)$; and for any $X, Y$ if $phi(X)$ and $phi(Y)$, then $X=Y$" is provable in ZFC (essentially, that the formula $phi$ determines a unique set). Then that set $X$ must belong to some level $L_alpha$ of the constructible hierarchy.
The argument is as follows: since the existence and uniqueness is provable in ZFC, it must be true in the constructible universe, which is a model of ZFC. But since constructible universe is an inner model of ZFC, that particular element of the constructible universe that serves as a witness of the truth of the formula in the constructible universe model of ZFC must also be the unique $X$ in the "whole" ZFC$-$we cannot have a different one, because there is already a set that satisfies $X$ in the constructible hierarchy.
Is this argument sound? It feels solid and completely obvious to me, and vacuous at the same time.
logic set-theory first-order-logic model-theory
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
asked Aug 10 at 21:50
xyzzyz
5,4611220
I was thinking about constructible universe, and I had the following idea.
Suppose we have a predicate $phi$ defined in a language of set theory. Suppose moreover, that the statement "there exists $X$ such that $phi(X)$; and for any $X, Y$ if $phi(X)$ and $phi(Y)$, then $X=Y$" is provable in ZFC (essentially, that the formula $phi$ determines a unique set). Then that set $X$ must belong to some level $L_alpha$ of the constructible hierarchy.
The argument is as follows: since the existence and uniqueness is provable in ZFC, it must be true in the constructible universe, which is a model of ZFC. But since constructible universe is an inner model of ZFC, that particular element of the constructible universe that serves as a witness of the truth of the formula in the constructible universe model of ZFC must also be the unique $X$ in the "whole" ZFC$-$we cannot have a different one, because there is already a set that satisfies $X$ in the constructible hierarchy.
Is this argument sound? It feels solid and completely obvious to me, and vacuous at the same time.
logic set-theory first-order-logic model-theory
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
asked Aug 10 at 21:50
xyzzyz
5,4611220
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
edited Aug 12 at 8:39
Taroccoesbrocco
3,72651433
3,72651433
asked Aug 10 at 21:50
xyzzyz
5,4611220
asked Aug 10 at 21:50
xyzzyz
5,4611220
asked Aug 10 at 21:50
xyzzyz
5,4611220
5,4611220
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
The argument is clearly wrong. $mathcal P(omega)$ is definable without parameters, and yet it is not necessarily constructible, as shown by Cohen.
The thing to remember is that the formula $phi$ is not absolute between models. So $mathcal P(omega)$ and $mathcal P(omega)^L$ might be different.
In some cases, there are sets whose definition is very robust and unique, but they cannot even exist in $L$. One example of this kind is $0^#$, which has a parameter free definition, and can be represented as a fairly canonical set of integers. But nevertheless, this set cannot exist in $L$. Other examples would be any real, since you can code it into the continuum pattern below $aleph_omega$ (for example), and even much more than that.
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
2
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
The argument is clearly wrong. $mathcal P(omega)$ is definable without parameters, and yet it is not necessarily constructible, as shown by Cohen.
The thing to remember is that the formula $phi$ is not absolute between models. So $mathcal P(omega)$ and $mathcal P(omega)^L$ might be different.
In some cases, there are sets whose definition is very robust and unique, but they cannot even exist in $L$. One example of this kind is $0^#$, which has a parameter free definition, and can be represented as a fairly canonical set of integers. But nevertheless, this set cannot exist in $L$. Other examples would be any real, since you can code it into the continuum pattern below $aleph_omega$ (for example), and even much more than that.
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
2
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
add a comment |Â
up vote
8
down vote
accepted
The argument is clearly wrong. $mathcal P(omega)$ is definable without parameters, and yet it is not necessarily constructible, as shown by Cohen.
The thing to remember is that the formula $phi$ is not absolute between models. So $mathcal P(omega)$ and $mathcal P(omega)^L$ might be different.
In some cases, there are sets whose definition is very robust and unique, but they cannot even exist in $L$. One example of this kind is $0^#$, which has a parameter free definition, and can be represented as a fairly canonical set of integers. But nevertheless, this set cannot exist in $L$. Other examples would be any real, since you can code it into the continuum pattern below $aleph_omega$ (for example), and even much more than that.
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
2
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
The argument is clearly wrong. $mathcal P(omega)$ is definable without parameters, and yet it is not necessarily constructible, as shown by Cohen.
The thing to remember is that the formula $phi$ is not absolute between models. So $mathcal P(omega)$ and $mathcal P(omega)^L$ might be different.
In some cases, there are sets whose definition is very robust and unique, but they cannot even exist in $L$. One example of this kind is $0^#$, which has a parameter free definition, and can be represented as a fairly canonical set of integers. But nevertheless, this set cannot exist in $L$. Other examples would be any real, since you can code it into the continuum pattern below $aleph_omega$ (for example), and even much more than that.
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
The argument is clearly wrong. $mathcal P(omega)$ is definable without parameters, and yet it is not necessarily constructible, as shown by Cohen.
The thing to remember is that the formula $phi$ is not absolute between models. So $mathcal P(omega)$ and $mathcal P(omega)^L$ might be different.
In some cases, there are sets whose definition is very robust and unique, but they cannot even exist in $L$. One example of this kind is $0^#$, which has a parameter free definition, and can be represented as a fairly canonical set of integers. But nevertheless, this set cannot exist in $L$. Other examples would be any real, since you can code it into the continuum pattern below $aleph_omega$ (for example), and even much more than that.
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
answered Aug 10 at 21:54
Asaf Karagila♦
294k31409736
294k31409736
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
2
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
add a comment |Â
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
2
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
Ah, so the point here is that even though $P(omega)$ is uniquely defined without parameters, then even though we might have $omega = omega^L$, we have that $P(omega)^L$ is the set of constructible subsets of $omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention?
– xyzzyz
Aug 10 at 22:24
2
2
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $Vneq L$ which has non-constructible reals.
– Asaf Karagila♦
Aug 10 at 22:25
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
Thank you! I'll look for some source on forcing.
– xyzzyz
Aug 12 at 4:18
add a comment |Â
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