Mixing
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Can we have a Choice-AntiChoice chain?
$begingroup$
Can we have a $kappa$ sized sequence $mathcal S$ of transitive domains $mathcal M_i$ of models of $``text{ZF + negation of choice}"$, where $kappa$ is inaccessible, such that for all $i < kappa$ we have $mathcal M_i in mathcal M_{i+1}$ and such that for each $mathcal M_i$ choice over$mathcal M_i$ applies in $mathcal M_{i+1}$ , i.e., for each element $s$ of $mathcal M_i$ there exists a set $c in mathcal M_{i+1}$ such that $c$ has exactly one element from each element of $s$ among its elements.
set-theory first-order-logic axiom-of-choice
asked Jan 23 at 18:11
ZuhairZuhair
340212
$endgroup$
add a comment |
$begingroup$
Can we have a $kappa$ sized sequence $mathcal S$ of transitive domains $mathcal M_i$ of models of $``text{ZF + negation of choice}"$, where $kappa$ is inaccessible, such that for all $i < kappa$ we have $mathcal M_i in mathcal M_{i+1}$ and such that for each $mathcal M_i$ choice over$mathcal M_i$ applies in $mathcal M_{i+1}$ , i.e., for each element $s$ of $mathcal M_i$ there exists a set $c in mathcal M_{i+1}$ such that $c$ has exactly one element from each element of $s$ among its elements.
set-theory first-order-logic axiom-of-choice
asked Jan 23 at 18:11
ZuhairZuhair
340212
$endgroup$
$begingroup$
Doesn't the naive approach work? Take $kappa$ of sufficient cardinal strength (e.g. $kappa$ Mahlo) and recursively build your sequence: At stage $i+1$, first collapse $mathcal{M_i}$ to be countable (which immediately gives that choice holds) and then extend it one step further to violate choice. Wrap all of this into a sufficiently large collapse of $V$ and you're done.
$endgroup$
– Stefan Mesken
Jan 24 at 13:07
$begingroup$
can you present that with some detail as an answer?
$endgroup$
– Zuhair
Jan 24 at 18:02
$begingroup$
Do you care about $kappa$'s cardinality or only its order-type? Because in the construction I've outlined above, $kappa$ will become countable and the question seems to become quite a bit harder if you want to preserve that $kappa$ is inaccessible. (The existence of suitable generic filters, in $V$, becomes problematic.)
$endgroup$
– Stefan Mesken
Jan 24 at 18:32
$begingroup$
well the idea is that $kappa$ is inaccessible.
$endgroup$
– Zuhair
Jan 24 at 18:42
$begingroup$
Okay, so this won't work (at least not without modification). I've translated your question in my head to 'can we have arbitrarily long choice/anti-choice sequences?' and only when typing the answer did I realize that you may have had something different in mind.
$endgroup$
– Stefan Mesken
Jan 24 at 18:48
add a comment |
$begingroup$
Can we have a $kappa$ sized sequence $mathcal S$ of transitive domains $mathcal M_i$ of models of $``text{ZF + negation of choice}"$, where $kappa$ is inaccessible, such that for all $i < kappa$ we have $mathcal M_i in mathcal M_{i+1}$ and such that for each $mathcal M_i$ choice over$mathcal M_i$ applies in $mathcal M_{i+1}$ , i.e., for each element $s$ of $mathcal M_i$ there exists a set $c in mathcal M_{i+1}$ such that $c$ has exactly one element from each element of $s$ among its elements.
set-theory first-order-logic axiom-of-choice
asked Jan 23 at 18:11
ZuhairZuhair
340212
$endgroup$
Can we have a $kappa$ sized sequence $mathcal S$ of transitive domains $mathcal M_i$ of models of $``text{ZF + negation of choice}"$, where $kappa$ is inaccessible, such that for all $i < kappa$ we have $mathcal M_i in mathcal M_{i+1}$ and such that for each $mathcal M_i$ choice over$mathcal M_i$ applies in $mathcal M_{i+1}$ , i.e., for each element $s$ of $mathcal M_i$ there exists a set $c in mathcal M_{i+1}$ such that $c$ has exactly one element from each element of $s$ among its elements.
set-theory first-order-logic axiom-of-choice
set-theory first-order-logic axiom-of-choice
asked Jan 23 at 18:11
ZuhairZuhair
340212
asked Jan 23 at 18:11
ZuhairZuhair
340212
asked Jan 23 at 18:11
ZuhairZuhair
340212
asked Jan 23 at 18:11
ZuhairZuhair
340212
asked Jan 23 at 18:11
ZuhairZuhair
340212
340212
$begingroup$
Doesn't the naive approach work? Take $kappa$ of sufficient cardinal strength (e.g. $kappa$ Mahlo) and recursively build your sequence: At stage $i+1$, first collapse $mathcal{M_i}$ to be countable (which immediately gives that choice holds) and then extend it one step further to violate choice. Wrap all of this into a sufficiently large collapse of $V$ and you're done.
$endgroup$
– Stefan Mesken
Jan 24 at 13:07
$begingroup$
can you present that with some detail as an answer?
$endgroup$
– Zuhair
Jan 24 at 18:02
$begingroup$
Do you care about $kappa$'s cardinality or only its order-type? Because in the construction I've outlined above, $kappa$ will become countable and the question seems to become quite a bit harder if you want to preserve that $kappa$ is inaccessible. (The existence of suitable generic filters, in $V$, becomes problematic.)
$endgroup$
– Stefan Mesken
Jan 24 at 18:32
$begingroup$
well the idea is that $kappa$ is inaccessible.
$endgroup$
– Zuhair
Jan 24 at 18:42
$begingroup$
Okay, so this won't work (at least not without modification). I've translated your question in my head to 'can we have arbitrarily long choice/anti-choice sequences?' and only when typing the answer did I realize that you may have had something different in mind.
$endgroup$
– Stefan Mesken
Jan 24 at 18:48
add a comment |
$begingroup$
Doesn't the naive approach work? Take $kappa$ of sufficient cardinal strength (e.g. $kappa$ Mahlo) and recursively build your sequence: At stage $i+1$, first collapse $mathcal{M_i}$ to be countable (which immediately gives that choice holds) and then extend it one step further to violate choice. Wrap all of this into a sufficiently large collapse of $V$ and you're done.
$endgroup$
– Stefan Mesken
Jan 24 at 13:07
$begingroup$
can you present that with some detail as an answer?
$endgroup$
– Zuhair
Jan 24 at 18:02
$begingroup$
Do you care about $kappa$'s cardinality or only its order-type? Because in the construction I've outlined above, $kappa$ will become countable and the question seems to become quite a bit harder if you want to preserve that $kappa$ is inaccessible. (The existence of suitable generic filters, in $V$, becomes problematic.)
$endgroup$
– Stefan Mesken
Jan 24 at 18:32
$begingroup$
well the idea is that $kappa$ is inaccessible.
$endgroup$
– Zuhair
Jan 24 at 18:42
$begingroup$
Okay, so this won't work (at least not without modification). I've translated your question in my head to 'can we have arbitrarily long choice/anti-choice sequences?' and only when typing the answer did I realize that you may have had something different in mind.
$endgroup$
– Stefan Mesken
Jan 24 at 18:48
$begingroup$
Doesn't the naive approach work? Take $kappa$ of sufficient cardinal strength (e.g. $kappa$ Mahlo) and recursively build your sequence: At stage $i+1$, first collapse $mathcal{M_i}$ to be countable (which immediately gives that choice holds) and then extend it one step further to violate choice. Wrap all of this into a sufficiently large collapse of $V$ and you're done.
$endgroup$
– Stefan Mesken
Jan 24 at 13:07
$begingroup$
Doesn't the naive approach work? Take $kappa$ of sufficient cardinal strength (e.g. $kappa$ Mahlo) and recursively build your sequence: At stage $i+1$, first collapse $mathcal{M_i}$ to be countable (which immediately gives that choice holds) and then extend it one step further to violate choice. Wrap all of this into a sufficiently large collapse of $V$ and you're done.
$endgroup$
– Stefan Mesken
Jan 24 at 13:07
$begingroup$
can you present that with some detail as an answer?
$endgroup$
– Zuhair
Jan 24 at 18:02
$begingroup$
can you present that with some detail as an answer?
$endgroup$
– Zuhair
Jan 24 at 18:02
$begingroup$
Do you care about $kappa$'s cardinality or only its order-type? Because in the construction I've outlined above, $kappa$ will become countable and the question seems to become quite a bit harder if you want to preserve that $kappa$ is inaccessible. (The existence of suitable generic filters, in $V$, becomes problematic.)
$endgroup$
– Stefan Mesken
Jan 24 at 18:32
$begingroup$
Do you care about $kappa$'s cardinality or only its order-type? Because in the construction I've outlined above, $kappa$ will become countable and the question seems to become quite a bit harder if you want to preserve that $kappa$ is inaccessible. (The existence of suitable generic filters, in $V$, becomes problematic.)
$endgroup$
– Stefan Mesken
Jan 24 at 18:32
$begingroup$
well the idea is that $kappa$ is inaccessible.
$endgroup$
– Zuhair
Jan 24 at 18:42
$begingroup$
well the idea is that $kappa$ is inaccessible.
$endgroup$
– Zuhair
Jan 24 at 18:42
$begingroup$
Okay, so this won't work (at least not without modification). I've translated your question in my head to 'can we have arbitrarily long choice/anti-choice sequences?' and only when typing the answer did I realize that you may have had something different in mind.
$endgroup$
– Stefan Mesken
Jan 24 at 18:48
$begingroup$
Okay, so this won't work (at least not without modification). I've translated your question in my head to 'can we have arbitrarily long choice/anti-choice sequences?' and only when typing the answer did I realize that you may have had something different in mind.
$endgroup$
– Stefan Mesken
Jan 24 at 18:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I believe the following works (and it's not even an iterated forcing argument):
Fix $kappa$ inaccessible . For a worldly cardinal $mu<kappa$, let $mathbb{P}_mu$ be the forcing consisting of maps from $mu$ to $2$ with domain of cardinality $<mu$. Let $(mu_eta)_{eta<kappa}$ enumerate the worldly cardinals $<kappa$ in increasing order (recalling that every inaccessible cardinal is a limit of worldly cardinals), and consider the forcing notion $$mathbb{P}={Pinprod_{eta<kappa}mathbb{P}_{mu_eta}: vert{alpha<kappa: P(alpha)not=emptyset}vert<kappa}$$ be the $<kappa$-support product of the $mathbb{P}_mu$s. Note that forcing with $mathbb{P}$ preserves the inaccessibility of $kappa$ (think about how $mathbb{P}$ "splits" into a $mu$-c.c. and a $mu$-closed piece, for every $mu<kappa$, and count the nice names ...).
Let $G$ be $mathbb{P}$-generic. For $eta<kappa$, let $G_eta=Gupharpoonright mu_eta$, and note that $G_eta$ is set-generic over the ZFC-model $L_{mu_{eta+1}}$. Now by a standard symmetric submodel argument we can whip up for each $eta$ a $W_eta$ such that:
$L_{mu_{eta+2}}subset W_etasubset L_{mu_{eta+2}}[G_{eta+1}],$
$W_etamodels$ ZF+$neg$AC, and
$G_etain W_eta$.
The $W_eta$s satisfy the desired properties (the third condition is what ensures that each one contains a well-ordering of the earlier ones).
EDIT: Deleted a broken idea.
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
$endgroup$
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe the following works (and it's not even an iterated forcing argument):
Fix $kappa$ inaccessible . For a worldly cardinal $mu<kappa$, let $mathbb{P}_mu$ be the forcing consisting of maps from $mu$ to $2$ with domain of cardinality $<mu$. Let $(mu_eta)_{eta<kappa}$ enumerate the worldly cardinals $<kappa$ in increasing order (recalling that every inaccessible cardinal is a limit of worldly cardinals), and consider the forcing notion $$mathbb{P}={Pinprod_{eta<kappa}mathbb{P}_{mu_eta}: vert{alpha<kappa: P(alpha)not=emptyset}vert<kappa}$$ be the $<kappa$-support product of the $mathbb{P}_mu$s. Note that forcing with $mathbb{P}$ preserves the inaccessibility of $kappa$ (think about how $mathbb{P}$ "splits" into a $mu$-c.c. and a $mu$-closed piece, for every $mu<kappa$, and count the nice names ...).
Let $G$ be $mathbb{P}$-generic. For $eta<kappa$, let $G_eta=Gupharpoonright mu_eta$, and note that $G_eta$ is set-generic over the ZFC-model $L_{mu_{eta+1}}$. Now by a standard symmetric submodel argument we can whip up for each $eta$ a $W_eta$ such that:
$L_{mu_{eta+2}}subset W_etasubset L_{mu_{eta+2}}[G_{eta+1}],$
$W_etamodels$ ZF+$neg$AC, and
$G_etain W_eta$.
The $W_eta$s satisfy the desired properties (the third condition is what ensures that each one contains a well-ordering of the earlier ones).
EDIT: Deleted a broken idea.
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
$endgroup$
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
add a comment |
$begingroup$
I believe the following works (and it's not even an iterated forcing argument):
Fix $kappa$ inaccessible . For a worldly cardinal $mu<kappa$, let $mathbb{P}_mu$ be the forcing consisting of maps from $mu$ to $2$ with domain of cardinality $<mu$. Let $(mu_eta)_{eta<kappa}$ enumerate the worldly cardinals $<kappa$ in increasing order (recalling that every inaccessible cardinal is a limit of worldly cardinals), and consider the forcing notion $$mathbb{P}={Pinprod_{eta<kappa}mathbb{P}_{mu_eta}: vert{alpha<kappa: P(alpha)not=emptyset}vert<kappa}$$ be the $<kappa$-support product of the $mathbb{P}_mu$s. Note that forcing with $mathbb{P}$ preserves the inaccessibility of $kappa$ (think about how $mathbb{P}$ "splits" into a $mu$-c.c. and a $mu$-closed piece, for every $mu<kappa$, and count the nice names ...).
Let $G$ be $mathbb{P}$-generic. For $eta<kappa$, let $G_eta=Gupharpoonright mu_eta$, and note that $G_eta$ is set-generic over the ZFC-model $L_{mu_{eta+1}}$. Now by a standard symmetric submodel argument we can whip up for each $eta$ a $W_eta$ such that:
$L_{mu_{eta+2}}subset W_etasubset L_{mu_{eta+2}}[G_{eta+1}],$
$W_etamodels$ ZF+$neg$AC, and
$G_etain W_eta$.
The $W_eta$s satisfy the desired properties (the third condition is what ensures that each one contains a well-ordering of the earlier ones).
EDIT: Deleted a broken idea.
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
$endgroup$
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
add a comment |
$begingroup$
I believe the following works (and it's not even an iterated forcing argument):
Fix $kappa$ inaccessible . For a worldly cardinal $mu<kappa$, let $mathbb{P}_mu$ be the forcing consisting of maps from $mu$ to $2$ with domain of cardinality $<mu$. Let $(mu_eta)_{eta<kappa}$ enumerate the worldly cardinals $<kappa$ in increasing order (recalling that every inaccessible cardinal is a limit of worldly cardinals), and consider the forcing notion $$mathbb{P}={Pinprod_{eta<kappa}mathbb{P}_{mu_eta}: vert{alpha<kappa: P(alpha)not=emptyset}vert<kappa}$$ be the $<kappa$-support product of the $mathbb{P}_mu$s. Note that forcing with $mathbb{P}$ preserves the inaccessibility of $kappa$ (think about how $mathbb{P}$ "splits" into a $mu$-c.c. and a $mu$-closed piece, for every $mu<kappa$, and count the nice names ...).
Let $G$ be $mathbb{P}$-generic. For $eta<kappa$, let $G_eta=Gupharpoonright mu_eta$, and note that $G_eta$ is set-generic over the ZFC-model $L_{mu_{eta+1}}$. Now by a standard symmetric submodel argument we can whip up for each $eta$ a $W_eta$ such that:
$L_{mu_{eta+2}}subset W_etasubset L_{mu_{eta+2}}[G_{eta+1}],$
$W_etamodels$ ZF+$neg$AC, and
$G_etain W_eta$.
The $W_eta$s satisfy the desired properties (the third condition is what ensures that each one contains a well-ordering of the earlier ones).
EDIT: Deleted a broken idea.
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
$endgroup$
I believe the following works (and it's not even an iterated forcing argument):
Fix $kappa$ inaccessible . For a worldly cardinal $mu<kappa$, let $mathbb{P}_mu$ be the forcing consisting of maps from $mu$ to $2$ with domain of cardinality $<mu$. Let $(mu_eta)_{eta<kappa}$ enumerate the worldly cardinals $<kappa$ in increasing order (recalling that every inaccessible cardinal is a limit of worldly cardinals), and consider the forcing notion $$mathbb{P}={Pinprod_{eta<kappa}mathbb{P}_{mu_eta}: vert{alpha<kappa: P(alpha)not=emptyset}vert<kappa}$$ be the $<kappa$-support product of the $mathbb{P}_mu$s. Note that forcing with $mathbb{P}$ preserves the inaccessibility of $kappa$ (think about how $mathbb{P}$ "splits" into a $mu$-c.c. and a $mu$-closed piece, for every $mu<kappa$, and count the nice names ...).
Let $G$ be $mathbb{P}$-generic. For $eta<kappa$, let $G_eta=Gupharpoonright mu_eta$, and note that $G_eta$ is set-generic over the ZFC-model $L_{mu_{eta+1}}$. Now by a standard symmetric submodel argument we can whip up for each $eta$ a $W_eta$ such that:
$L_{mu_{eta+2}}subset W_etasubset L_{mu_{eta+2}}[G_{eta+1}],$
$W_etamodels$ ZF+$neg$AC, and
$G_etain W_eta$.
The $W_eta$s satisfy the desired properties (the third condition is what ensures that each one contains a well-ordering of the earlier ones).
EDIT: Deleted a broken idea.
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
edited Feb 8 at 21:47
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
answered Feb 8 at 18:41
Noah SchweberNoah Schweber
127k10151290
127k10151290
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
add a comment |
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
The second part is false. If $eta=beth_delta+1$ where $delta$ has uncountable cofinality, then $V_etacongmathcal P(omega_delta)$ and by a theorem of Shelah, $L(mathcal P(kappa))$ satisfies choice when $kappa$ is singular of uncountable cofinality.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:23
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
@AsafKaragila Oh that's fascinating! I'd never seen that before. Clearly my memory is off - do you know if it holds for successor-of-successor $eta$s? (And do you have a citation for Shelah's result?)
$endgroup$
– Noah Schweber
Feb 8 at 21:47
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
$begingroup$
Not off hand, I'll try to find one tomorrow or so. I think it's in the choiceless PCF papers, maybe Shelah 835.
$endgroup$
– Asaf Karagila♦
Feb 8 at 21:57
add a comment |
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Doesn't the naive approach work? Take $kappa$ of sufficient cardinal strength (e.g. $kappa$ Mahlo) and recursively build your sequence: At stage $i+1$, first collapse $mathcal{M_i}$ to be countable (which immediately gives that choice holds) and then extend it one step further to violate choice. Wrap all of this into a sufficiently large collapse of $V$ and you're done.
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– Stefan Mesken
Jan 24 at 13:07
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can you present that with some detail as an answer?
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– Zuhair
Jan 24 at 18:02
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Do you care about $kappa$'s cardinality or only its order-type? Because in the construction I've outlined above, $kappa$ will become countable and the question seems to become quite a bit harder if you want to preserve that $kappa$ is inaccessible. (The existence of suitable generic filters, in $V$, becomes problematic.)
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– Stefan Mesken
Jan 24 at 18:32
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well the idea is that $kappa$ is inaccessible.
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– Zuhair
Jan 24 at 18:42
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Okay, so this won't work (at least not without modification). I've translated your question in my head to 'can we have arbitrarily long choice/anti-choice sequences?' and only when typing the answer did I realize that you may have had something different in mind.
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– Stefan Mesken
Jan 24 at 18:48