Existence of disjunction of conditions in countable support iteration of Laver forcing
$begingroup$
The part i'm interested in, is the part between the brackets. One side note is that the forcing order is reversed and that's why the disjunction is the greatest lower bound instead of least upper bound. My confusion here is the part which i have underlined in the image. How do we take the conjunction of such $q Rightarrow p_q(gamma)$? My first guess is to embed $mathbb{P}^{alpha beta}$ into a complete boolean algebra and go from there. But this idea has an apparent issue and that is the fact that $q not in mathbb{P}^{alpha beta}$ and we can't calculate $q Rightarrow p_q(gamma)$. What is he doing here? Because the disjunction is crucial in the proofs of his last lemmas, i can't skip it. Any help would be appreciated.
Edit I: One idea i have is is to outright ignore that $Rightarrow$ and go with this: Let $p(gamma)$ be a name for the union of the $p_q(gamma)$ such that the names for the $p_q(gamma)$ are altered so that when a generic set is chosen containing another $q'$ from the antichain like $G$ then $p_q(gamma)$ is empty. This does seem to meet the requirements by writing them down. Is this correct? If so, can we extract the meaning of $Rightarrow$ from this?(Because it appears again in the bottom of the page.)
set-theory forcing
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
$endgroup$
|
show 2 more comments
$begingroup$
The part i'm interested in, is the part between the brackets. One side note is that the forcing order is reversed and that's why the disjunction is the greatest lower bound instead of least upper bound. My confusion here is the part which i have underlined in the image. How do we take the conjunction of such $q Rightarrow p_q(gamma)$? My first guess is to embed $mathbb{P}^{alpha beta}$ into a complete boolean algebra and go from there. But this idea has an apparent issue and that is the fact that $q not in mathbb{P}^{alpha beta}$ and we can't calculate $q Rightarrow p_q(gamma)$. What is he doing here? Because the disjunction is crucial in the proofs of his last lemmas, i can't skip it. Any help would be appreciated.
Edit I: One idea i have is is to outright ignore that $Rightarrow$ and go with this: Let $p(gamma)$ be a name for the union of the $p_q(gamma)$ such that the names for the $p_q(gamma)$ are altered so that when a generic set is chosen containing another $q'$ from the antichain like $G$ then $p_q(gamma)$ is empty. This does seem to meet the requirements by writing them down. Is this correct? If so, can we extract the meaning of $Rightarrow$ from this?(Because it appears again in the bottom of the page.)
set-theory forcing
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
$endgroup$
$begingroup$
Any ideas on what that arrow could mean?
$endgroup$
– Shervin Sorouri
Dec 28 '18 at 11:30
1
$begingroup$
I think in 2018 this statement you are confused by would be written as $qVdash_alpha p(gamma) = p_q(gamma)$.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:32
$begingroup$
Oh, but it's still vague how someone could extract a condition from the conjunction of some forcing relations.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 7:50
2
$begingroup$
Just mix the names over the antichain.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:51
$begingroup$
But then again one thing that may cause trouble is the fact that if $gamma > alpha$ then the name for $p_q(gamma)$ isn't even in the language of $mathbb{P}_{alpha}$.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 8:00
|
show 2 more comments
$begingroup$
The part i'm interested in, is the part between the brackets. One side note is that the forcing order is reversed and that's why the disjunction is the greatest lower bound instead of least upper bound. My confusion here is the part which i have underlined in the image. How do we take the conjunction of such $q Rightarrow p_q(gamma)$? My first guess is to embed $mathbb{P}^{alpha beta}$ into a complete boolean algebra and go from there. But this idea has an apparent issue and that is the fact that $q not in mathbb{P}^{alpha beta}$ and we can't calculate $q Rightarrow p_q(gamma)$. What is he doing here? Because the disjunction is crucial in the proofs of his last lemmas, i can't skip it. Any help would be appreciated.
Edit I: One idea i have is is to outright ignore that $Rightarrow$ and go with this: Let $p(gamma)$ be a name for the union of the $p_q(gamma)$ such that the names for the $p_q(gamma)$ are altered so that when a generic set is chosen containing another $q'$ from the antichain like $G$ then $p_q(gamma)$ is empty. This does seem to meet the requirements by writing them down. Is this correct? If so, can we extract the meaning of $Rightarrow$ from this?(Because it appears again in the bottom of the page.)
set-theory forcing
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
$endgroup$
The part i'm interested in, is the part between the brackets. One side note is that the forcing order is reversed and that's why the disjunction is the greatest lower bound instead of least upper bound. My confusion here is the part which i have underlined in the image. How do we take the conjunction of such $q Rightarrow p_q(gamma)$? My first guess is to embed $mathbb{P}^{alpha beta}$ into a complete boolean algebra and go from there. But this idea has an apparent issue and that is the fact that $q not in mathbb{P}^{alpha beta}$ and we can't calculate $q Rightarrow p_q(gamma)$. What is he doing here? Because the disjunction is crucial in the proofs of his last lemmas, i can't skip it. Any help would be appreciated.
Edit I: One idea i have is is to outright ignore that $Rightarrow$ and go with this: Let $p(gamma)$ be a name for the union of the $p_q(gamma)$ such that the names for the $p_q(gamma)$ are altered so that when a generic set is chosen containing another $q'$ from the antichain like $G$ then $p_q(gamma)$ is empty. This does seem to meet the requirements by writing them down. Is this correct? If so, can we extract the meaning of $Rightarrow$ from this?(Because it appears again in the bottom of the page.)
set-theory forcing
set-theory forcing
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
edited Dec 29 '18 at 12:00
Asaf Karagila♦
308k33441775
308k33441775
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
asked Dec 27 '18 at 8:47
Shervin SorouriShervin Sorouri
519212
519212
$begingroup$
Any ideas on what that arrow could mean?
$endgroup$
– Shervin Sorouri
Dec 28 '18 at 11:30
1
$begingroup$
I think in 2018 this statement you are confused by would be written as $qVdash_alpha p(gamma) = p_q(gamma)$.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:32
$begingroup$
Oh, but it's still vague how someone could extract a condition from the conjunction of some forcing relations.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 7:50
2
$begingroup$
Just mix the names over the antichain.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:51
$begingroup$
But then again one thing that may cause trouble is the fact that if $gamma > alpha$ then the name for $p_q(gamma)$ isn't even in the language of $mathbb{P}_{alpha}$.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 8:00
|
show 2 more comments
$begingroup$
Any ideas on what that arrow could mean?
$endgroup$
– Shervin Sorouri
Dec 28 '18 at 11:30
1
$begingroup$
I think in 2018 this statement you are confused by would be written as $qVdash_alpha p(gamma) = p_q(gamma)$.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:32
$begingroup$
Oh, but it's still vague how someone could extract a condition from the conjunction of some forcing relations.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 7:50
2
$begingroup$
Just mix the names over the antichain.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:51
$begingroup$
But then again one thing that may cause trouble is the fact that if $gamma > alpha$ then the name for $p_q(gamma)$ isn't even in the language of $mathbb{P}_{alpha}$.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 8:00
$begingroup$
Any ideas on what that arrow could mean?
$endgroup$
– Shervin Sorouri
Dec 28 '18 at 11:30
$begingroup$
Any ideas on what that arrow could mean?
$endgroup$
– Shervin Sorouri
Dec 28 '18 at 11:30
1
1
$begingroup$
I think in 2018 this statement you are confused by would be written as $qVdash_alpha p(gamma) = p_q(gamma)$.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:32
$begingroup$
I think in 2018 this statement you are confused by would be written as $qVdash_alpha p(gamma) = p_q(gamma)$.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:32
$begingroup$
Oh, but it's still vague how someone could extract a condition from the conjunction of some forcing relations.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 7:50
$begingroup$
Oh, but it's still vague how someone could extract a condition from the conjunction of some forcing relations.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 7:50
2
2
$begingroup$
Just mix the names over the antichain.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:51
$begingroup$
Just mix the names over the antichain.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:51
$begingroup$
But then again one thing that may cause trouble is the fact that if $gamma > alpha$ then the name for $p_q(gamma)$ isn't even in the language of $mathbb{P}_{alpha}$.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 8:00
$begingroup$
But then again one thing that may cause trouble is the fact that if $gamma > alpha$ then the name for $p_q(gamma)$ isn't even in the language of $mathbb{P}_{alpha}$.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 8:00
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The point here is that when you iterate, you can amalgamate a bunch of names of conditions in "future iterations" into a single condition in the full length iteration.
Namely, given a bunch of conditions in $Bbb P_beta$, whose projections to $Bbb P_alpha$ all extend some fixed $q'$, we have a condition in $Bbb P_beta$ whose projection to $Bbb P_alpha$ is $q'$, and the content of condition in $Bbb{P_beta/P_alpha}$ is decided by some extension of $q'$ in $Bbb P_alpha$.
To be more formal, this means that given $q'$ we take a maximal antichain below it in $Bbb P_alpha$, and to each $q$ in the antichain we attach a condition $p_qinBbb P_beta$ such that $p_qrestrictionalpha=q$, and then we define a condition $p$ in $Bbb P_beta$ such that $prestrictionalpha=q'$, and for each $q$ in the antichain, if $q$ is in the generic for $Bbb P_alpha$, then $p$ is interpreted the same as $p_q$.
In principle it just means that $Bbb{P_betacong P_alphaast P_beta/P_alpha}$. The key point of Baumgartner is that this amount of freedom does not exist in a product, since in a product we require the conditions of the iteration to be canonical ground model names, rather than arbitrary names which allow for some genericity to be incorporated into the partial order.
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053719%2fexistence-of-disjunction-of-conditions-in-countable-support-iteration-of-laver-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The point here is that when you iterate, you can amalgamate a bunch of names of conditions in "future iterations" into a single condition in the full length iteration.
Namely, given a bunch of conditions in $Bbb P_beta$, whose projections to $Bbb P_alpha$ all extend some fixed $q'$, we have a condition in $Bbb P_beta$ whose projection to $Bbb P_alpha$ is $q'$, and the content of condition in $Bbb{P_beta/P_alpha}$ is decided by some extension of $q'$ in $Bbb P_alpha$.
To be more formal, this means that given $q'$ we take a maximal antichain below it in $Bbb P_alpha$, and to each $q$ in the antichain we attach a condition $p_qinBbb P_beta$ such that $p_qrestrictionalpha=q$, and then we define a condition $p$ in $Bbb P_beta$ such that $prestrictionalpha=q'$, and for each $q$ in the antichain, if $q$ is in the generic for $Bbb P_alpha$, then $p$ is interpreted the same as $p_q$.
In principle it just means that $Bbb{P_betacong P_alphaast P_beta/P_alpha}$. The key point of Baumgartner is that this amount of freedom does not exist in a product, since in a product we require the conditions of the iteration to be canonical ground model names, rather than arbitrary names which allow for some genericity to be incorporated into the partial order.
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
The point here is that when you iterate, you can amalgamate a bunch of names of conditions in "future iterations" into a single condition in the full length iteration.
Namely, given a bunch of conditions in $Bbb P_beta$, whose projections to $Bbb P_alpha$ all extend some fixed $q'$, we have a condition in $Bbb P_beta$ whose projection to $Bbb P_alpha$ is $q'$, and the content of condition in $Bbb{P_beta/P_alpha}$ is decided by some extension of $q'$ in $Bbb P_alpha$.
To be more formal, this means that given $q'$ we take a maximal antichain below it in $Bbb P_alpha$, and to each $q$ in the antichain we attach a condition $p_qinBbb P_beta$ such that $p_qrestrictionalpha=q$, and then we define a condition $p$ in $Bbb P_beta$ such that $prestrictionalpha=q'$, and for each $q$ in the antichain, if $q$ is in the generic for $Bbb P_alpha$, then $p$ is interpreted the same as $p_q$.
In principle it just means that $Bbb{P_betacong P_alphaast P_beta/P_alpha}$. The key point of Baumgartner is that this amount of freedom does not exist in a product, since in a product we require the conditions of the iteration to be canonical ground model names, rather than arbitrary names which allow for some genericity to be incorporated into the partial order.
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
The point here is that when you iterate, you can amalgamate a bunch of names of conditions in "future iterations" into a single condition in the full length iteration.
Namely, given a bunch of conditions in $Bbb P_beta$, whose projections to $Bbb P_alpha$ all extend some fixed $q'$, we have a condition in $Bbb P_beta$ whose projection to $Bbb P_alpha$ is $q'$, and the content of condition in $Bbb{P_beta/P_alpha}$ is decided by some extension of $q'$ in $Bbb P_alpha$.
To be more formal, this means that given $q'$ we take a maximal antichain below it in $Bbb P_alpha$, and to each $q$ in the antichain we attach a condition $p_qinBbb P_beta$ such that $p_qrestrictionalpha=q$, and then we define a condition $p$ in $Bbb P_beta$ such that $prestrictionalpha=q'$, and for each $q$ in the antichain, if $q$ is in the generic for $Bbb P_alpha$, then $p$ is interpreted the same as $p_q$.
In principle it just means that $Bbb{P_betacong P_alphaast P_beta/P_alpha}$. The key point of Baumgartner is that this amount of freedom does not exist in a product, since in a product we require the conditions of the iteration to be canonical ground model names, rather than arbitrary names which allow for some genericity to be incorporated into the partial order.
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
The point here is that when you iterate, you can amalgamate a bunch of names of conditions in "future iterations" into a single condition in the full length iteration.
Namely, given a bunch of conditions in $Bbb P_beta$, whose projections to $Bbb P_alpha$ all extend some fixed $q'$, we have a condition in $Bbb P_beta$ whose projection to $Bbb P_alpha$ is $q'$, and the content of condition in $Bbb{P_beta/P_alpha}$ is decided by some extension of $q'$ in $Bbb P_alpha$.
To be more formal, this means that given $q'$ we take a maximal antichain below it in $Bbb P_alpha$, and to each $q$ in the antichain we attach a condition $p_qinBbb P_beta$ such that $p_qrestrictionalpha=q$, and then we define a condition $p$ in $Bbb P_beta$ such that $prestrictionalpha=q'$, and for each $q$ in the antichain, if $q$ is in the generic for $Bbb P_alpha$, then $p$ is interpreted the same as $p_q$.
In principle it just means that $Bbb{P_betacong P_alphaast P_beta/P_alpha}$. The key point of Baumgartner is that this amount of freedom does not exist in a product, since in a product we require the conditions of the iteration to be canonical ground model names, rather than arbitrary names which allow for some genericity to be incorporated into the partial order.
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
answered Dec 29 '18 at 11:58
Asaf Karagila♦Asaf Karagila
308k33441775
308k33441775
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053719%2fexistence-of-disjunction-of-conditions-in-countable-support-iteration-of-laver-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Any ideas on what that arrow could mean?
$endgroup$
– Shervin Sorouri
Dec 28 '18 at 11:30
1
$begingroup$
I think in 2018 this statement you are confused by would be written as $qVdash_alpha p(gamma) = p_q(gamma)$.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:32
$begingroup$
Oh, but it's still vague how someone could extract a condition from the conjunction of some forcing relations.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 7:50
2
$begingroup$
Just mix the names over the antichain.
$endgroup$
– Asaf Karagila♦
Dec 29 '18 at 7:51
$begingroup$
But then again one thing that may cause trouble is the fact that if $gamma > alpha$ then the name for $p_q(gamma)$ isn't even in the language of $mathbb{P}_{alpha}$.
$endgroup$
– Shervin Sorouri
Dec 29 '18 at 8:00