Minimal model of ZF with $0sharp$
We know that the constructible universe $L$ is an absolute and minimal model of ZF (every standard model of ZF contains "an" $L$, and it is actually the same $L$ for all of them).
It is also my understanding that the existence of $0sharp$ informally means that $V$ is much "bigger" than $L$ (meaning that if $0sharp$ exists then even $aleph_1$ is already an inaccessible cardinal in $L$) and that a sort of converse is true (see: https://en.wikipedia.org/wiki/Jensen%27s_covering_theorem).
Therefore my question is: Is there an absolute minimal model of ZF + $exists0sharp$ ?
A sort of "$Lsharp$" if we really want to abuse notation?
logic set-theory model-theory large-cardinals
|
show 1 more comment
We know that the constructible universe $L$ is an absolute and minimal model of ZF (every standard model of ZF contains "an" $L$, and it is actually the same $L$ for all of them).
It is also my understanding that the existence of $0sharp$ informally means that $V$ is much "bigger" than $L$ (meaning that if $0sharp$ exists then even $aleph_1$ is already an inaccessible cardinal in $L$) and that a sort of converse is true (see: https://en.wikipedia.org/wiki/Jensen%27s_covering_theorem).
Therefore my question is: Is there an absolute minimal model of ZF + $exists0sharp$ ?
A sort of "$Lsharp$" if we really want to abuse notation?
logic set-theory model-theory large-cardinals
4
Sure. The model is denoted $L[0^sharp]$. You mimic the construction of $L$, but allow in your language a predicate for $0^sharp$ (understood as a set of numbers).
– Andrés E. Caicedo
Feb 28 '16 at 0:18
Isn't $L(0sharp)$ smaller?
– Alon Navon
Feb 28 '16 at 0:26
4
We have $L[0^sharp] = L(0^sharp)$: In general, for any given set $A$, we have $L[A] subseteq L(A)$ with equalitiy if and only if $A cap L[A] = A$. Since $0^sharp$ may be regarded as a subset of $omega$, we have $0^sharp subseteq L subseteq L[0^sharp]$ and thus the claimed equality.
– Stefan Mesken
Feb 28 '16 at 0:38
1
@Stefan Thank you very much, I got my relative constructibility all mixed up, and you cleared a lot. :) This means in essence that we can define a "sharp" sequence. $L_0 = L$, $L_1 = L[0^sharp]$, $L_2 = L[0^{sharpsharp}]$, etc... Just wondering whether there is any use for this? Any other nice properties that make these "L's" "L-ish"?
– Alon Navon
Feb 28 '16 at 0:49
3
Well, $L[A]$ is similar to $L$ in a lot of ways - a major reason for this is, that the usual condensation lemma generalizes to $L[A]$ (which allows us to prove $operatorname{GCH}$, $Diamond_kappa$, $square_kappa$, ... on a "tail segment" of $L[A]$).
– Stefan Mesken
Feb 28 '16 at 1:19
|
show 1 more comment
We know that the constructible universe $L$ is an absolute and minimal model of ZF (every standard model of ZF contains "an" $L$, and it is actually the same $L$ for all of them).
It is also my understanding that the existence of $0sharp$ informally means that $V$ is much "bigger" than $L$ (meaning that if $0sharp$ exists then even $aleph_1$ is already an inaccessible cardinal in $L$) and that a sort of converse is true (see: https://en.wikipedia.org/wiki/Jensen%27s_covering_theorem).
Therefore my question is: Is there an absolute minimal model of ZF + $exists0sharp$ ?
A sort of "$Lsharp$" if we really want to abuse notation?
logic set-theory model-theory large-cardinals
We know that the constructible universe $L$ is an absolute and minimal model of ZF (every standard model of ZF contains "an" $L$, and it is actually the same $L$ for all of them).
It is also my understanding that the existence of $0sharp$ informally means that $V$ is much "bigger" than $L$ (meaning that if $0sharp$ exists then even $aleph_1$ is already an inaccessible cardinal in $L$) and that a sort of converse is true (see: https://en.wikipedia.org/wiki/Jensen%27s_covering_theorem).
Therefore my question is: Is there an absolute minimal model of ZF + $exists0sharp$ ?
A sort of "$Lsharp$" if we really want to abuse notation?
logic set-theory model-theory large-cardinals
logic set-theory model-theory large-cardinals
asked Feb 28 '16 at 0:12
Alon Navon
1,025413
1,025413
4
Sure. The model is denoted $L[0^sharp]$. You mimic the construction of $L$, but allow in your language a predicate for $0^sharp$ (understood as a set of numbers).
– Andrés E. Caicedo
Feb 28 '16 at 0:18
Isn't $L(0sharp)$ smaller?
– Alon Navon
Feb 28 '16 at 0:26
4
We have $L[0^sharp] = L(0^sharp)$: In general, for any given set $A$, we have $L[A] subseteq L(A)$ with equalitiy if and only if $A cap L[A] = A$. Since $0^sharp$ may be regarded as a subset of $omega$, we have $0^sharp subseteq L subseteq L[0^sharp]$ and thus the claimed equality.
– Stefan Mesken
Feb 28 '16 at 0:38
1
@Stefan Thank you very much, I got my relative constructibility all mixed up, and you cleared a lot. :) This means in essence that we can define a "sharp" sequence. $L_0 = L$, $L_1 = L[0^sharp]$, $L_2 = L[0^{sharpsharp}]$, etc... Just wondering whether there is any use for this? Any other nice properties that make these "L's" "L-ish"?
– Alon Navon
Feb 28 '16 at 0:49
3
Well, $L[A]$ is similar to $L$ in a lot of ways - a major reason for this is, that the usual condensation lemma generalizes to $L[A]$ (which allows us to prove $operatorname{GCH}$, $Diamond_kappa$, $square_kappa$, ... on a "tail segment" of $L[A]$).
– Stefan Mesken
Feb 28 '16 at 1:19
|
show 1 more comment
4
Sure. The model is denoted $L[0^sharp]$. You mimic the construction of $L$, but allow in your language a predicate for $0^sharp$ (understood as a set of numbers).
– Andrés E. Caicedo
Feb 28 '16 at 0:18
Isn't $L(0sharp)$ smaller?
– Alon Navon
Feb 28 '16 at 0:26
4
We have $L[0^sharp] = L(0^sharp)$: In general, for any given set $A$, we have $L[A] subseteq L(A)$ with equalitiy if and only if $A cap L[A] = A$. Since $0^sharp$ may be regarded as a subset of $omega$, we have $0^sharp subseteq L subseteq L[0^sharp]$ and thus the claimed equality.
– Stefan Mesken
Feb 28 '16 at 0:38
1
@Stefan Thank you very much, I got my relative constructibility all mixed up, and you cleared a lot. :) This means in essence that we can define a "sharp" sequence. $L_0 = L$, $L_1 = L[0^sharp]$, $L_2 = L[0^{sharpsharp}]$, etc... Just wondering whether there is any use for this? Any other nice properties that make these "L's" "L-ish"?
– Alon Navon
Feb 28 '16 at 0:49
3
Well, $L[A]$ is similar to $L$ in a lot of ways - a major reason for this is, that the usual condensation lemma generalizes to $L[A]$ (which allows us to prove $operatorname{GCH}$, $Diamond_kappa$, $square_kappa$, ... on a "tail segment" of $L[A]$).
– Stefan Mesken
Feb 28 '16 at 1:19
4
4
Sure. The model is denoted $L[0^sharp]$. You mimic the construction of $L$, but allow in your language a predicate for $0^sharp$ (understood as a set of numbers).
– Andrés E. Caicedo
Feb 28 '16 at 0:18
Sure. The model is denoted $L[0^sharp]$. You mimic the construction of $L$, but allow in your language a predicate for $0^sharp$ (understood as a set of numbers).
– Andrés E. Caicedo
Feb 28 '16 at 0:18
Isn't $L(0sharp)$ smaller?
– Alon Navon
Feb 28 '16 at 0:26
Isn't $L(0sharp)$ smaller?
– Alon Navon
Feb 28 '16 at 0:26
4
4
We have $L[0^sharp] = L(0^sharp)$: In general, for any given set $A$, we have $L[A] subseteq L(A)$ with equalitiy if and only if $A cap L[A] = A$. Since $0^sharp$ may be regarded as a subset of $omega$, we have $0^sharp subseteq L subseteq L[0^sharp]$ and thus the claimed equality.
– Stefan Mesken
Feb 28 '16 at 0:38
We have $L[0^sharp] = L(0^sharp)$: In general, for any given set $A$, we have $L[A] subseteq L(A)$ with equalitiy if and only if $A cap L[A] = A$. Since $0^sharp$ may be regarded as a subset of $omega$, we have $0^sharp subseteq L subseteq L[0^sharp]$ and thus the claimed equality.
– Stefan Mesken
Feb 28 '16 at 0:38
1
1
@Stefan Thank you very much, I got my relative constructibility all mixed up, and you cleared a lot. :) This means in essence that we can define a "sharp" sequence. $L_0 = L$, $L_1 = L[0^sharp]$, $L_2 = L[0^{sharpsharp}]$, etc... Just wondering whether there is any use for this? Any other nice properties that make these "L's" "L-ish"?
– Alon Navon
Feb 28 '16 at 0:49
@Stefan Thank you very much, I got my relative constructibility all mixed up, and you cleared a lot. :) This means in essence that we can define a "sharp" sequence. $L_0 = L$, $L_1 = L[0^sharp]$, $L_2 = L[0^{sharpsharp}]$, etc... Just wondering whether there is any use for this? Any other nice properties that make these "L's" "L-ish"?
– Alon Navon
Feb 28 '16 at 0:49
3
3
Well, $L[A]$ is similar to $L$ in a lot of ways - a major reason for this is, that the usual condensation lemma generalizes to $L[A]$ (which allows us to prove $operatorname{GCH}$, $Diamond_kappa$, $square_kappa$, ... on a "tail segment" of $L[A]$).
– Stefan Mesken
Feb 28 '16 at 1:19
Well, $L[A]$ is similar to $L$ in a lot of ways - a major reason for this is, that the usual condensation lemma generalizes to $L[A]$ (which allows us to prove $operatorname{GCH}$, $Diamond_kappa$, $square_kappa$, ... on a "tail segment" of $L[A]$).
– Stefan Mesken
Feb 28 '16 at 1:19
|
show 1 more comment
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Yes, the minimal such model is $L[0^sharp]$. This model can be built by stages, just as $L$, starting with the empty set, taking unions at limit stages, and at each successor stage $alpha+1$ taking the collection of subsets of $L_alpha[0^sharp]$ definable in $(L_alpha[0^sharp],in,0^sharp)$ from parameters. Here, $0^sharp$ can be thought of as a set of natural numbers, and definability is in the language of set theory with one additional predicate.
Note that for each finite $n$, $L_n[0^sharp]=L_n$ (that is, the universes of both structures coincide) and so $L_omega[0^sharp]=L_omega$. However, $0^sharp$ is now definable at this stage (as the set of numbers satisfying the new predicate), so $0^sharpin L[0^sharp]$.
This is not quite enough. Let $varphi(x)$ be the formula in the language of set theory stating that $x$ is $0^sharp$, i.e., stating that $x$ is the unique EM blueprint satisfying the three indiscernibility conditions listed in section 9 of
MR1994835 (2004f:03092) Kanamori, Akihiro. The higher infinite. Large cardinals in set theory from their beginnings. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xxii+536 pp. ISBN: 3-540-00384-3.
One should also check that $L[0^sharp]models varphi(0^sharp)$, and that whenever $M$ is an inner model and $Mmodelsvarphi(a)$ for some $a$, then $0^sharp$ exists in $V$ and $a=0^sharp$, so that $L[0^sharp]subseteq M$. This is essentially an absoluteness argument, but it is a bit technical so I will skip the details here. The point of proving this is that not only is $L[0^sharp]$ the smallest such model, but it knows it, in the sense that it satisfies $L[0^sharp]models V=L[0^sharp]$, and is contained in any inner model that believes in the existence of $0^sharp$.
One can prove more as well, for instance, $L[0^sharp]$ satisfies appropriate analogues of the fine structural properties of $L$, so it is not just the least model containing $0^sharp$, it is also a very well-behaved model. Naturally, the construction generalizes to more general sharps and other inner-model theoretic objects, although the absoluteness requirements become more involved.
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Yes, the minimal such model is $L[0^sharp]$. This model can be built by stages, just as $L$, starting with the empty set, taking unions at limit stages, and at each successor stage $alpha+1$ taking the collection of subsets of $L_alpha[0^sharp]$ definable in $(L_alpha[0^sharp],in,0^sharp)$ from parameters. Here, $0^sharp$ can be thought of as a set of natural numbers, and definability is in the language of set theory with one additional predicate.
Note that for each finite $n$, $L_n[0^sharp]=L_n$ (that is, the universes of both structures coincide) and so $L_omega[0^sharp]=L_omega$. However, $0^sharp$ is now definable at this stage (as the set of numbers satisfying the new predicate), so $0^sharpin L[0^sharp]$.
This is not quite enough. Let $varphi(x)$ be the formula in the language of set theory stating that $x$ is $0^sharp$, i.e., stating that $x$ is the unique EM blueprint satisfying the three indiscernibility conditions listed in section 9 of
MR1994835 (2004f:03092) Kanamori, Akihiro. The higher infinite. Large cardinals in set theory from their beginnings. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xxii+536 pp. ISBN: 3-540-00384-3.
One should also check that $L[0^sharp]models varphi(0^sharp)$, and that whenever $M$ is an inner model and $Mmodelsvarphi(a)$ for some $a$, then $0^sharp$ exists in $V$ and $a=0^sharp$, so that $L[0^sharp]subseteq M$. This is essentially an absoluteness argument, but it is a bit technical so I will skip the details here. The point of proving this is that not only is $L[0^sharp]$ the smallest such model, but it knows it, in the sense that it satisfies $L[0^sharp]models V=L[0^sharp]$, and is contained in any inner model that believes in the existence of $0^sharp$.
One can prove more as well, for instance, $L[0^sharp]$ satisfies appropriate analogues of the fine structural properties of $L$, so it is not just the least model containing $0^sharp$, it is also a very well-behaved model. Naturally, the construction generalizes to more general sharps and other inner-model theoretic objects, although the absoluteness requirements become more involved.
add a comment |
Yes, the minimal such model is $L[0^sharp]$. This model can be built by stages, just as $L$, starting with the empty set, taking unions at limit stages, and at each successor stage $alpha+1$ taking the collection of subsets of $L_alpha[0^sharp]$ definable in $(L_alpha[0^sharp],in,0^sharp)$ from parameters. Here, $0^sharp$ can be thought of as a set of natural numbers, and definability is in the language of set theory with one additional predicate.
Note that for each finite $n$, $L_n[0^sharp]=L_n$ (that is, the universes of both structures coincide) and so $L_omega[0^sharp]=L_omega$. However, $0^sharp$ is now definable at this stage (as the set of numbers satisfying the new predicate), so $0^sharpin L[0^sharp]$.
This is not quite enough. Let $varphi(x)$ be the formula in the language of set theory stating that $x$ is $0^sharp$, i.e., stating that $x$ is the unique EM blueprint satisfying the three indiscernibility conditions listed in section 9 of
MR1994835 (2004f:03092) Kanamori, Akihiro. The higher infinite. Large cardinals in set theory from their beginnings. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xxii+536 pp. ISBN: 3-540-00384-3.
One should also check that $L[0^sharp]models varphi(0^sharp)$, and that whenever $M$ is an inner model and $Mmodelsvarphi(a)$ for some $a$, then $0^sharp$ exists in $V$ and $a=0^sharp$, so that $L[0^sharp]subseteq M$. This is essentially an absoluteness argument, but it is a bit technical so I will skip the details here. The point of proving this is that not only is $L[0^sharp]$ the smallest such model, but it knows it, in the sense that it satisfies $L[0^sharp]models V=L[0^sharp]$, and is contained in any inner model that believes in the existence of $0^sharp$.
One can prove more as well, for instance, $L[0^sharp]$ satisfies appropriate analogues of the fine structural properties of $L$, so it is not just the least model containing $0^sharp$, it is also a very well-behaved model. Naturally, the construction generalizes to more general sharps and other inner-model theoretic objects, although the absoluteness requirements become more involved.
add a comment |
Yes, the minimal such model is $L[0^sharp]$. This model can be built by stages, just as $L$, starting with the empty set, taking unions at limit stages, and at each successor stage $alpha+1$ taking the collection of subsets of $L_alpha[0^sharp]$ definable in $(L_alpha[0^sharp],in,0^sharp)$ from parameters. Here, $0^sharp$ can be thought of as a set of natural numbers, and definability is in the language of set theory with one additional predicate.
Note that for each finite $n$, $L_n[0^sharp]=L_n$ (that is, the universes of both structures coincide) and so $L_omega[0^sharp]=L_omega$. However, $0^sharp$ is now definable at this stage (as the set of numbers satisfying the new predicate), so $0^sharpin L[0^sharp]$.
This is not quite enough. Let $varphi(x)$ be the formula in the language of set theory stating that $x$ is $0^sharp$, i.e., stating that $x$ is the unique EM blueprint satisfying the three indiscernibility conditions listed in section 9 of
MR1994835 (2004f:03092) Kanamori, Akihiro. The higher infinite. Large cardinals in set theory from their beginnings. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xxii+536 pp. ISBN: 3-540-00384-3.
One should also check that $L[0^sharp]models varphi(0^sharp)$, and that whenever $M$ is an inner model and $Mmodelsvarphi(a)$ for some $a$, then $0^sharp$ exists in $V$ and $a=0^sharp$, so that $L[0^sharp]subseteq M$. This is essentially an absoluteness argument, but it is a bit technical so I will skip the details here. The point of proving this is that not only is $L[0^sharp]$ the smallest such model, but it knows it, in the sense that it satisfies $L[0^sharp]models V=L[0^sharp]$, and is contained in any inner model that believes in the existence of $0^sharp$.
One can prove more as well, for instance, $L[0^sharp]$ satisfies appropriate analogues of the fine structural properties of $L$, so it is not just the least model containing $0^sharp$, it is also a very well-behaved model. Naturally, the construction generalizes to more general sharps and other inner-model theoretic objects, although the absoluteness requirements become more involved.
Yes, the minimal such model is $L[0^sharp]$. This model can be built by stages, just as $L$, starting with the empty set, taking unions at limit stages, and at each successor stage $alpha+1$ taking the collection of subsets of $L_alpha[0^sharp]$ definable in $(L_alpha[0^sharp],in,0^sharp)$ from parameters. Here, $0^sharp$ can be thought of as a set of natural numbers, and definability is in the language of set theory with one additional predicate.
Note that for each finite $n$, $L_n[0^sharp]=L_n$ (that is, the universes of both structures coincide) and so $L_omega[0^sharp]=L_omega$. However, $0^sharp$ is now definable at this stage (as the set of numbers satisfying the new predicate), so $0^sharpin L[0^sharp]$.
This is not quite enough. Let $varphi(x)$ be the formula in the language of set theory stating that $x$ is $0^sharp$, i.e., stating that $x$ is the unique EM blueprint satisfying the three indiscernibility conditions listed in section 9 of
MR1994835 (2004f:03092) Kanamori, Akihiro. The higher infinite. Large cardinals in set theory from their beginnings. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xxii+536 pp. ISBN: 3-540-00384-3.
One should also check that $L[0^sharp]models varphi(0^sharp)$, and that whenever $M$ is an inner model and $Mmodelsvarphi(a)$ for some $a$, then $0^sharp$ exists in $V$ and $a=0^sharp$, so that $L[0^sharp]subseteq M$. This is essentially an absoluteness argument, but it is a bit technical so I will skip the details here. The point of proving this is that not only is $L[0^sharp]$ the smallest such model, but it knows it, in the sense that it satisfies $L[0^sharp]models V=L[0^sharp]$, and is contained in any inner model that believes in the existence of $0^sharp$.
One can prove more as well, for instance, $L[0^sharp]$ satisfies appropriate analogues of the fine structural properties of $L$, so it is not just the least model containing $0^sharp$, it is also a very well-behaved model. Naturally, the construction generalizes to more general sharps and other inner-model theoretic objects, although the absoluteness requirements become more involved.
answered Nov 27 '18 at 18:32
Andrés E. Caicedo
64.7k8158246
64.7k8158246
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4
Sure. The model is denoted $L[0^sharp]$. You mimic the construction of $L$, but allow in your language a predicate for $0^sharp$ (understood as a set of numbers).
– Andrés E. Caicedo
Feb 28 '16 at 0:18
Isn't $L(0sharp)$ smaller?
– Alon Navon
Feb 28 '16 at 0:26
4
We have $L[0^sharp] = L(0^sharp)$: In general, for any given set $A$, we have $L[A] subseteq L(A)$ with equalitiy if and only if $A cap L[A] = A$. Since $0^sharp$ may be regarded as a subset of $omega$, we have $0^sharp subseteq L subseteq L[0^sharp]$ and thus the claimed equality.
– Stefan Mesken
Feb 28 '16 at 0:38
1
@Stefan Thank you very much, I got my relative constructibility all mixed up, and you cleared a lot. :) This means in essence that we can define a "sharp" sequence. $L_0 = L$, $L_1 = L[0^sharp]$, $L_2 = L[0^{sharpsharp}]$, etc... Just wondering whether there is any use for this? Any other nice properties that make these "L's" "L-ish"?
– Alon Navon
Feb 28 '16 at 0:49
3
Well, $L[A]$ is similar to $L$ in a lot of ways - a major reason for this is, that the usual condensation lemma generalizes to $L[A]$ (which allows us to prove $operatorname{GCH}$, $Diamond_kappa$, $square_kappa$, ... on a "tail segment" of $L[A]$).
– Stefan Mesken
Feb 28 '16 at 1:19