Standard fact about iterated forcing
$begingroup$
I'm studying Laver's proof of the consistency of Borel's conjecture and he says that the following is a standard fact about iterated forcing:
Define $mathbb{P}^{alpha beta}$ as the set of all functions $f$ with domain $[alpha, beta)$ such that $1_{alpha} cup f in mathbb{P}_{beta}$ and order $mathbb{P}^{alpha beta}$ in $mathfrak{M}[G_{alpha}]$ with:
$f le g longleftrightarrow (exists p in G_{alpha}) p cup f le pcup g$
Let $G_{beta}$ be generic over $mathbb{P}_{beta}$, then $mathfrak{M}[G_{beta}] = mathfrak{M}[G_{alpha}][G^{alpha beta}]$ where $G^{alpha beta}$ is $mathfrak{M}[G_{alpha}]text{-generic}$ over $mathbb{P}^{alpha beta}$.
My thoughts:
I tried to prove $G_{beta} = G_{alpha} * G^{alpha beta}$ and that $G^{alpha beta} = G^{beta}|_{[alpha, beta)}$ but i failed because $G^{alpha beta}$ is so random. I would be really glad if someone pointed me in the right direction.
Edit I: The definitions that i forgot:
$G_{alpha} = G_{beta} |_{alpha}$ and $1_{alpha}$ is the canonical name for the greatest element of $mathbb{P}_{alpha}$ and $mathbb{P}_{beta}$ is the $betatext{-th}$ stage iteration of Laver forcing.
set-theory forcing
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
$endgroup$
add a comment |
$begingroup$
I'm studying Laver's proof of the consistency of Borel's conjecture and he says that the following is a standard fact about iterated forcing:
Define $mathbb{P}^{alpha beta}$ as the set of all functions $f$ with domain $[alpha, beta)$ such that $1_{alpha} cup f in mathbb{P}_{beta}$ and order $mathbb{P}^{alpha beta}$ in $mathfrak{M}[G_{alpha}]$ with:
$f le g longleftrightarrow (exists p in G_{alpha}) p cup f le pcup g$
Let $G_{beta}$ be generic over $mathbb{P}_{beta}$, then $mathfrak{M}[G_{beta}] = mathfrak{M}[G_{alpha}][G^{alpha beta}]$ where $G^{alpha beta}$ is $mathfrak{M}[G_{alpha}]text{-generic}$ over $mathbb{P}^{alpha beta}$.
My thoughts:
I tried to prove $G_{beta} = G_{alpha} * G^{alpha beta}$ and that $G^{alpha beta} = G^{beta}|_{[alpha, beta)}$ but i failed because $G^{alpha beta}$ is so random. I would be really glad if someone pointed me in the right direction.
Edit I: The definitions that i forgot:
$G_{alpha} = G_{beta} |_{alpha}$ and $1_{alpha}$ is the canonical name for the greatest element of $mathbb{P}_{alpha}$ and $mathbb{P}_{beta}$ is the $betatext{-th}$ stage iteration of Laver forcing.
set-theory forcing
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
$endgroup$
3
$begingroup$
This is a fairly standard fact for iterations. But I agree, it can be tricky to figure out how to prove these things when you're just starting with forcing. The easiest way is to define the partial order of $Bbb P_alphaastBbb P^{alphabeta}$ and show it is isomorphic to $Bbb P_beta$, and that should do most of the job for you. The rest is done by noting that the isomorphism is in fact the identity on $Bbb P_alphaast{1_{alphabeta}}$.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:24
1
$begingroup$
Oh right then by the two-step iteration lemma we are done, right?
$endgroup$
– Shervin Sorouri
Dec 11 '18 at 11:30
2
$begingroup$
Yes, pretty much.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:30
add a comment |
$begingroup$
I'm studying Laver's proof of the consistency of Borel's conjecture and he says that the following is a standard fact about iterated forcing:
Define $mathbb{P}^{alpha beta}$ as the set of all functions $f$ with domain $[alpha, beta)$ such that $1_{alpha} cup f in mathbb{P}_{beta}$ and order $mathbb{P}^{alpha beta}$ in $mathfrak{M}[G_{alpha}]$ with:
$f le g longleftrightarrow (exists p in G_{alpha}) p cup f le pcup g$
Let $G_{beta}$ be generic over $mathbb{P}_{beta}$, then $mathfrak{M}[G_{beta}] = mathfrak{M}[G_{alpha}][G^{alpha beta}]$ where $G^{alpha beta}$ is $mathfrak{M}[G_{alpha}]text{-generic}$ over $mathbb{P}^{alpha beta}$.
My thoughts:
I tried to prove $G_{beta} = G_{alpha} * G^{alpha beta}$ and that $G^{alpha beta} = G^{beta}|_{[alpha, beta)}$ but i failed because $G^{alpha beta}$ is so random. I would be really glad if someone pointed me in the right direction.
Edit I: The definitions that i forgot:
$G_{alpha} = G_{beta} |_{alpha}$ and $1_{alpha}$ is the canonical name for the greatest element of $mathbb{P}_{alpha}$ and $mathbb{P}_{beta}$ is the $betatext{-th}$ stage iteration of Laver forcing.
set-theory forcing
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
$endgroup$
I'm studying Laver's proof of the consistency of Borel's conjecture and he says that the following is a standard fact about iterated forcing:
Define $mathbb{P}^{alpha beta}$ as the set of all functions $f$ with domain $[alpha, beta)$ such that $1_{alpha} cup f in mathbb{P}_{beta}$ and order $mathbb{P}^{alpha beta}$ in $mathfrak{M}[G_{alpha}]$ with:
$f le g longleftrightarrow (exists p in G_{alpha}) p cup f le pcup g$
Let $G_{beta}$ be generic over $mathbb{P}_{beta}$, then $mathfrak{M}[G_{beta}] = mathfrak{M}[G_{alpha}][G^{alpha beta}]$ where $G^{alpha beta}$ is $mathfrak{M}[G_{alpha}]text{-generic}$ over $mathbb{P}^{alpha beta}$.
My thoughts:
I tried to prove $G_{beta} = G_{alpha} * G^{alpha beta}$ and that $G^{alpha beta} = G^{beta}|_{[alpha, beta)}$ but i failed because $G^{alpha beta}$ is so random. I would be really glad if someone pointed me in the right direction.
Edit I: The definitions that i forgot:
$G_{alpha} = G_{beta} |_{alpha}$ and $1_{alpha}$ is the canonical name for the greatest element of $mathbb{P}_{alpha}$ and $mathbb{P}_{beta}$ is the $betatext{-th}$ stage iteration of Laver forcing.
set-theory forcing
set-theory forcing
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
edited Dec 12 '18 at 14:26
Asaf Karagila♦
305k33435765
305k33435765
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
asked Dec 11 '18 at 10:40
Shervin SorouriShervin Sorouri
519211
519211
3
$begingroup$
This is a fairly standard fact for iterations. But I agree, it can be tricky to figure out how to prove these things when you're just starting with forcing. The easiest way is to define the partial order of $Bbb P_alphaastBbb P^{alphabeta}$ and show it is isomorphic to $Bbb P_beta$, and that should do most of the job for you. The rest is done by noting that the isomorphism is in fact the identity on $Bbb P_alphaast{1_{alphabeta}}$.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:24
1
$begingroup$
Oh right then by the two-step iteration lemma we are done, right?
$endgroup$
– Shervin Sorouri
Dec 11 '18 at 11:30
2
$begingroup$
Yes, pretty much.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:30
add a comment |
3
$begingroup$
This is a fairly standard fact for iterations. But I agree, it can be tricky to figure out how to prove these things when you're just starting with forcing. The easiest way is to define the partial order of $Bbb P_alphaastBbb P^{alphabeta}$ and show it is isomorphic to $Bbb P_beta$, and that should do most of the job for you. The rest is done by noting that the isomorphism is in fact the identity on $Bbb P_alphaast{1_{alphabeta}}$.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:24
1
$begingroup$
Oh right then by the two-step iteration lemma we are done, right?
$endgroup$
– Shervin Sorouri
Dec 11 '18 at 11:30
2
$begingroup$
Yes, pretty much.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:30
3
3
$begingroup$
This is a fairly standard fact for iterations. But I agree, it can be tricky to figure out how to prove these things when you're just starting with forcing. The easiest way is to define the partial order of $Bbb P_alphaastBbb P^{alphabeta}$ and show it is isomorphic to $Bbb P_beta$, and that should do most of the job for you. The rest is done by noting that the isomorphism is in fact the identity on $Bbb P_alphaast{1_{alphabeta}}$.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:24
$begingroup$
This is a fairly standard fact for iterations. But I agree, it can be tricky to figure out how to prove these things when you're just starting with forcing. The easiest way is to define the partial order of $Bbb P_alphaastBbb P^{alphabeta}$ and show it is isomorphic to $Bbb P_beta$, and that should do most of the job for you. The rest is done by noting that the isomorphism is in fact the identity on $Bbb P_alphaast{1_{alphabeta}}$.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:24
1
1
$begingroup$
Oh right then by the two-step iteration lemma we are done, right?
$endgroup$
– Shervin Sorouri
Dec 11 '18 at 11:30
$begingroup$
Oh right then by the two-step iteration lemma we are done, right?
$endgroup$
– Shervin Sorouri
Dec 11 '18 at 11:30
2
2
$begingroup$
Yes, pretty much.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:30
$begingroup$
Yes, pretty much.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:30
add a comment |
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$begingroup$
This is a fairly standard fact for iterations. But I agree, it can be tricky to figure out how to prove these things when you're just starting with forcing. The easiest way is to define the partial order of $Bbb P_alphaastBbb P^{alphabeta}$ and show it is isomorphic to $Bbb P_beta$, and that should do most of the job for you. The rest is done by noting that the isomorphism is in fact the identity on $Bbb P_alphaast{1_{alphabeta}}$.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:24
1
$begingroup$
Oh right then by the two-step iteration lemma we are done, right?
$endgroup$
– Shervin Sorouri
Dec 11 '18 at 11:30
2
$begingroup$
Yes, pretty much.
$endgroup$
– Asaf Karagila♦
Dec 11 '18 at 11:30