Minimal model of ZF with $0sharp$












3














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?










share|cite|improve this question


















  • 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


















3














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?










share|cite|improve this question


















  • 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
















3












3








3


1





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?










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












1 Answer
1






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oldest

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3














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.






    share|cite|improve this answer


























      3














      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.






      share|cite|improve this answer
























        3












        3








        3






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 18:32









        Andrés E. Caicedo

        64.7k8158246




        64.7k8158246






























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