Cohen forcing factoring
I start from $M$ a transitive countable model of $ZFC + mathbb V= mathbb L$ and I add a single Cohen generic $G$.
Now if $A in M[G]$ is also Cohen generic over $mathbb L$ and $M[A] ne M[G]$, can I deduce that there is some $G'$ Cohen generic over $M[A]$ such that $M[A][G']=M[G]$?
Thanks
logic set-theory forcing
asked Jun 24 '15 at 6:35
Régis
111
add a comment |
I start from $M$ a transitive countable model of $ZFC + mathbb V= mathbb L$ and I add a single Cohen generic $G$.
Now if $A in M[G]$ is also Cohen generic over $mathbb L$ and $M[A] ne M[G]$, can I deduce that there is some $G'$ Cohen generic over $M[A]$ such that $M[A][G']=M[G]$?
Thanks
logic set-theory forcing
asked Jun 24 '15 at 6:35
Régis
111
add a comment |
I start from $M$ a transitive countable model of $ZFC + mathbb V= mathbb L$ and I add a single Cohen generic $G$.
Now if $A in M[G]$ is also Cohen generic over $mathbb L$ and $M[A] ne M[G]$, can I deduce that there is some $G'$ Cohen generic over $M[A]$ such that $M[A][G']=M[G]$?
Thanks
logic set-theory forcing
asked Jun 24 '15 at 6:35
Régis
111
I start from $M$ a transitive countable model of $ZFC + mathbb V= mathbb L$ and I add a single Cohen generic $G$.
Now if $A in M[G]$ is also Cohen generic over $mathbb L$ and $M[A] ne M[G]$, can I deduce that there is some $G'$ Cohen generic over $M[A]$ such that $M[A][G']=M[G]$?
Thanks
logic set-theory forcing
logic set-theory forcing
asked Jun 24 '15 at 6:35
Régis
111
asked Jun 24 '15 at 6:35
Régis
111
asked Jun 24 '15 at 6:35
Régis
111
asked Jun 24 '15 at 6:35
Régis
111
asked Jun 24 '15 at 6:35
Régis
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
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The answer is yes. Recall the intermediate model theorem:
If $Msubseteq Nsubseteq M[G]$ are all models of $sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.
If you follow the proof, you will see that we construct quotients of the forcing used to construct $M[G]$. In the case of the Cohen forcing, a quotient is either atomic, or isomorphic to the Cohen forcing itself.
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
|
show 1 more comment
The full result sounds: If $M[a]$ is a Cohen extension of $M$ and $b$ is a real in $M[a]$ then one of the following three options takes place: 1) $bin M$, 2)$M[b]=M[a]$, 3) $M[b]$ is a Cohen extension of $M$ (not necessarily that $b$ itself is a Cohen real) and $M[a]$ is a Cohen extension of $M[b]$ (again not necessarily that $a$ itself is a Cohen real). An earliest mention of this is, afaik, in Ramez Sami thesis entitled Questions in descriptive set-theory and the determinacy of infinite games, Berkeley, 1976, where he refers to Vopenka-Hajek. A clean set-forcing (no BA stuff) proof is eg in my DOI: 10.17377/smzh.2017.58.610, where the key argument, obscured in the BA setting, is that both extensions are induced by countable forcing notions.
By the way the same result is true for random-forcing extensions, but the proof given in arXiv:1811.10568 is way more complex.
answered Nov 27 at 4:33
Vladimir Kanovei
1786
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The answer is yes. Recall the intermediate model theorem:
If $Msubseteq Nsubseteq M[G]$ are all models of $sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.
If you follow the proof, you will see that we construct quotients of the forcing used to construct $M[G]$. In the case of the Cohen forcing, a quotient is either atomic, or isomorphic to the Cohen forcing itself.
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
|
show 1 more comment
The answer is yes. Recall the intermediate model theorem:
If $Msubseteq Nsubseteq M[G]$ are all models of $sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.
If you follow the proof, you will see that we construct quotients of the forcing used to construct $M[G]$. In the case of the Cohen forcing, a quotient is either atomic, or isomorphic to the Cohen forcing itself.
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
|
show 1 more comment
The answer is yes. Recall the intermediate model theorem:
If $Msubseteq Nsubseteq M[G]$ are all models of $sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.
If you follow the proof, you will see that we construct quotients of the forcing used to construct $M[G]$. In the case of the Cohen forcing, a quotient is either atomic, or isomorphic to the Cohen forcing itself.
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
The answer is yes. Recall the intermediate model theorem:
If $Msubseteq Nsubseteq M[G]$ are all models of $sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.
If you follow the proof, you will see that we construct quotients of the forcing used to construct $M[G]$. In the case of the Cohen forcing, a quotient is either atomic, or isomorphic to the Cohen forcing itself.
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
edited Jun 25 '15 at 7:09
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
answered Jun 24 '15 at 7:42
Asaf Karagila♦
301k32424755
301k32424755
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
|
show 1 more comment
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Thank you very much, but where can I find this proof?
– Régis
Jun 24 '15 at 9:02
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
Jech, "Set Theory", I believe at the end of chapter 14.
– Asaf Karagila♦
Jun 24 '15 at 9:03
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
I still haven't gotten over how neat this theorem is . . . :)
– Noah Schweber
Jun 24 '15 at 14:07
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
@Noah: Quite. This theorem extends to $sf ZF$, although truth be told, I never fully sat down to understand the extension. You can find a discussion in Griegorieff's "Intermediate submodels and generic extensions in set theory" (Annals of Mathematics, vol. 101 no. 3)
– Asaf Karagila♦
Jun 24 '15 at 15:02
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
I may be wrong, but it seems that Griegorieff attributes this result to Solovay.
– Régis
Jun 25 '15 at 6:30
|
show 1 more comment
The full result sounds: If $M[a]$ is a Cohen extension of $M$ and $b$ is a real in $M[a]$ then one of the following three options takes place: 1) $bin M$, 2)$M[b]=M[a]$, 3) $M[b]$ is a Cohen extension of $M$ (not necessarily that $b$ itself is a Cohen real) and $M[a]$ is a Cohen extension of $M[b]$ (again not necessarily that $a$ itself is a Cohen real). An earliest mention of this is, afaik, in Ramez Sami thesis entitled Questions in descriptive set-theory and the determinacy of infinite games, Berkeley, 1976, where he refers to Vopenka-Hajek. A clean set-forcing (no BA stuff) proof is eg in my DOI: 10.17377/smzh.2017.58.610, where the key argument, obscured in the BA setting, is that both extensions are induced by countable forcing notions.
By the way the same result is true for random-forcing extensions, but the proof given in arXiv:1811.10568 is way more complex.
answered Nov 27 at 4:33
Vladimir Kanovei
1786
add a comment |
The full result sounds: If $M[a]$ is a Cohen extension of $M$ and $b$ is a real in $M[a]$ then one of the following three options takes place: 1) $bin M$, 2)$M[b]=M[a]$, 3) $M[b]$ is a Cohen extension of $M$ (not necessarily that $b$ itself is a Cohen real) and $M[a]$ is a Cohen extension of $M[b]$ (again not necessarily that $a$ itself is a Cohen real). An earliest mention of this is, afaik, in Ramez Sami thesis entitled Questions in descriptive set-theory and the determinacy of infinite games, Berkeley, 1976, where he refers to Vopenka-Hajek. A clean set-forcing (no BA stuff) proof is eg in my DOI: 10.17377/smzh.2017.58.610, where the key argument, obscured in the BA setting, is that both extensions are induced by countable forcing notions.
By the way the same result is true for random-forcing extensions, but the proof given in arXiv:1811.10568 is way more complex.
answered Nov 27 at 4:33
Vladimir Kanovei
1786
add a comment |
The full result sounds: If $M[a]$ is a Cohen extension of $M$ and $b$ is a real in $M[a]$ then one of the following three options takes place: 1) $bin M$, 2)$M[b]=M[a]$, 3) $M[b]$ is a Cohen extension of $M$ (not necessarily that $b$ itself is a Cohen real) and $M[a]$ is a Cohen extension of $M[b]$ (again not necessarily that $a$ itself is a Cohen real). An earliest mention of this is, afaik, in Ramez Sami thesis entitled Questions in descriptive set-theory and the determinacy of infinite games, Berkeley, 1976, where he refers to Vopenka-Hajek. A clean set-forcing (no BA stuff) proof is eg in my DOI: 10.17377/smzh.2017.58.610, where the key argument, obscured in the BA setting, is that both extensions are induced by countable forcing notions.
By the way the same result is true for random-forcing extensions, but the proof given in arXiv:1811.10568 is way more complex.
answered Nov 27 at 4:33
Vladimir Kanovei
1786
The full result sounds: If $M[a]$ is a Cohen extension of $M$ and $b$ is a real in $M[a]$ then one of the following three options takes place: 1) $bin M$, 2)$M[b]=M[a]$, 3) $M[b]$ is a Cohen extension of $M$ (not necessarily that $b$ itself is a Cohen real) and $M[a]$ is a Cohen extension of $M[b]$ (again not necessarily that $a$ itself is a Cohen real). An earliest mention of this is, afaik, in Ramez Sami thesis entitled Questions in descriptive set-theory and the determinacy of infinite games, Berkeley, 1976, where he refers to Vopenka-Hajek. A clean set-forcing (no BA stuff) proof is eg in my DOI: 10.17377/smzh.2017.58.610, where the key argument, obscured in the BA setting, is that both extensions are induced by countable forcing notions.
By the way the same result is true for random-forcing extensions, but the proof given in arXiv:1811.10568 is way more complex.
answered Nov 27 at 4:33
Vladimir Kanovei
1786
answered Nov 27 at 4:33
Vladimir Kanovei
1786
answered Nov 27 at 4:33
Vladimir Kanovei
1786
answered Nov 27 at 4:33
Vladimir Kanovei
1786
1786
add a comment |
add a comment |
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