Do we know the consistency strength of the Singular Cardinal Hypothesis failing on an uncountable cofinality?Disjoint stationary sets that reflectConsistency strength of the failure of Shelah's Strong Hypothesis (SSH)Prevalent singular cardinals hypothesisSlim Kurepa tree at a singular strong limit cardinal of uncountable cofinalitystationary tower forcingVery weak square and good pointsThe cofinality of the poset $[kappa]^<kappa$ for a singular cardinal $kappa$What are some good lower bounds on the consistency of the failure of the PCF conjecture?
Do we know the consistency strength of the Singular Cardinal Hypothesis failing on an uncountable cofinality?
Disjoint stationary sets that reflectConsistency strength of the failure of Shelah's Strong Hypothesis (SSH)Prevalent singular cardinals hypothesisSlim Kurepa tree at a singular strong limit cardinal of uncountable cofinalitystationary tower forcingVery weak square and good pointsThe cofinality of the poset $[kappa]^<kappa$ for a singular cardinal $kappa$What are some good lower bounds on the consistency of the failure of the PCF conjecture?
$begingroup$
Suppose that $kappa$ is a strong limit cardinal. The singular cardinal hypothesis states $2^kappa=kappa^+$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent with a measurable cardinal $kappa$ satisfying $o(kappa)=kappa^++$.
But this failure is at $aleph_omega$. Suppose we wanted more.
Suppose that we wanted the failure to happen on a couple isolated points. Well, it's not hard to redo the standard constructions and get just that. But what happens when we have limit points?
Even more, by Silver's theorem if SCH fails at $kappa>operatornamecf(kappa)>omega$, then there is a stationary subset of $kappa$ where SCH failed.
What would be the consistency strength when $kappa$ is a singular limit of singular cardinals, and SCH fails cofinally below $kappa$? What if we require $kappa$ to be of uncountable cofinality?
As a side question, what if $kappa$, with uncountable cofinality, does satisfy SCH, but an unbounded subset (which has to be non-stationary, of course) of it does not?
set-theory lo.logic large-cardinals
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose that $kappa$ is a strong limit cardinal. The singular cardinal hypothesis states $2^kappa=kappa^+$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent with a measurable cardinal $kappa$ satisfying $o(kappa)=kappa^++$.
But this failure is at $aleph_omega$. Suppose we wanted more.
Suppose that we wanted the failure to happen on a couple isolated points. Well, it's not hard to redo the standard constructions and get just that. But what happens when we have limit points?
Even more, by Silver's theorem if SCH fails at $kappa>operatornamecf(kappa)>omega$, then there is a stationary subset of $kappa$ where SCH failed.
What would be the consistency strength when $kappa$ is a singular limit of singular cardinals, and SCH fails cofinally below $kappa$? What if we require $kappa$ to be of uncountable cofinality?
As a side question, what if $kappa$, with uncountable cofinality, does satisfy SCH, but an unbounded subset (which has to be non-stationary, of course) of it does not?
set-theory lo.logic large-cardinals
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
$endgroup$
$begingroup$
Could you remind the set-theorist dilettantes what $o(kappa)$ stands for?
$endgroup$
– Gro-Tsen
Apr 15 at 10:27
3
$begingroup$
@Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(kappa)=1$ means just being a measurable, but $o(kappa)=2$ means that there is a normal measure which concentrates on those $o(lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank
$endgroup$
– Asaf Karagila
Apr 15 at 11:11
add a comment
|
$begingroup$
Suppose that $kappa$ is a strong limit cardinal. The singular cardinal hypothesis states $2^kappa=kappa^+$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent with a measurable cardinal $kappa$ satisfying $o(kappa)=kappa^++$.
But this failure is at $aleph_omega$. Suppose we wanted more.
Suppose that we wanted the failure to happen on a couple isolated points. Well, it's not hard to redo the standard constructions and get just that. But what happens when we have limit points?
Even more, by Silver's theorem if SCH fails at $kappa>operatornamecf(kappa)>omega$, then there is a stationary subset of $kappa$ where SCH failed.
What would be the consistency strength when $kappa$ is a singular limit of singular cardinals, and SCH fails cofinally below $kappa$? What if we require $kappa$ to be of uncountable cofinality?
As a side question, what if $kappa$, with uncountable cofinality, does satisfy SCH, but an unbounded subset (which has to be non-stationary, of course) of it does not?
set-theory lo.logic large-cardinals
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
$endgroup$
Suppose that $kappa$ is a strong limit cardinal. The singular cardinal hypothesis states $2^kappa=kappa^+$. We know that the failure of SCH requires large cardinals, and in fact is equiconsistent with a measurable cardinal $kappa$ satisfying $o(kappa)=kappa^++$.
But this failure is at $aleph_omega$. Suppose we wanted more.
Suppose that we wanted the failure to happen on a couple isolated points. Well, it's not hard to redo the standard constructions and get just that. But what happens when we have limit points?
Even more, by Silver's theorem if SCH fails at $kappa>operatornamecf(kappa)>omega$, then there is a stationary subset of $kappa$ where SCH failed.
What would be the consistency strength when $kappa$ is a singular limit of singular cardinals, and SCH fails cofinally below $kappa$? What if we require $kappa$ to be of uncountable cofinality?
As a side question, what if $kappa$, with uncountable cofinality, does satisfy SCH, but an unbounded subset (which has to be non-stationary, of course) of it does not?
set-theory lo.logic large-cardinals
set-theory lo.logic large-cardinals
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
asked Apr 15 at 8:16
Asaf KaragilaAsaf Karagila
22.4k6 gold badges83 silver badges192 bronze badges
22.4k6 gold badges83 silver badges192 bronze badges
$begingroup$
Could you remind the set-theorist dilettantes what $o(kappa)$ stands for?
$endgroup$
– Gro-Tsen
Apr 15 at 10:27
3
$begingroup$
@Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(kappa)=1$ means just being a measurable, but $o(kappa)=2$ means that there is a normal measure which concentrates on those $o(lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank
$endgroup$
– Asaf Karagila
Apr 15 at 11:11
add a comment
|
$begingroup$
Could you remind the set-theorist dilettantes what $o(kappa)$ stands for?
$endgroup$
– Gro-Tsen
Apr 15 at 10:27
3
$begingroup$
@Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(kappa)=1$ means just being a measurable, but $o(kappa)=2$ means that there is a normal measure which concentrates on those $o(lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank
$endgroup$
– Asaf Karagila
Apr 15 at 11:11
$begingroup$
Could you remind the set-theorist dilettantes what $o(kappa)$ stands for?
$endgroup$
– Gro-Tsen
Apr 15 at 10:27
$begingroup$
Could you remind the set-theorist dilettantes what $o(kappa)$ stands for?
$endgroup$
– Gro-Tsen
Apr 15 at 10:27
3
3
$begingroup$
@Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(kappa)=1$ means just being a measurable, but $o(kappa)=2$ means that there is a normal measure which concentrates on those $o(lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank
$endgroup$
– Asaf Karagila
Apr 15 at 11:11
$begingroup$
@Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(kappa)=1$ means just being a measurable, but $o(kappa)=2$ means that there is a normal measure which concentrates on those $o(lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank
$endgroup$
– Asaf Karagila
Apr 15 at 11:11
add a comment
|
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $kappa$ is a singular cardinal and there are $cf(kappa)$-many measurable cardinals $lambda < kappa$ with $o(lambda)=lambda^++$ cofinal in $kappa.$ Then you can perform a Prikry type iteration and get the failure of $SCH$ at cofinally many singular cardinals below $kappa.$
Now suppose we also want for $SCH$ to fail at $kappa$ itself. First let us consider the countable cofinality.
Assume $kappa$ is a measurable cardinal with $o(kappa)=kappa^+++1.$ Then we can get an extension in which $cf(kappa)=omega, 2^kappa=kappa^++$ and for cofinally many singular cardinals $lambda$ below $kappa$, we have $2^lambda=lambda^++$. I don't know if this assumption is really needed or if it can be reduced.
For uncountable cofinality, say $theta$, a measurable cardinal $kappa$ with $o(kappa)=kappa^+++theta$ is sufficient. As then you can first find an extension in which $2^kappa=kappa^++$ and such that in the extension, $o(kappa)=theta$. Then if you perform Magidor forcing for changing cofinality of $kappa$ to $theta,$ you can get a club $C$ of singular cardinals below $kappa$ such that for all $lambda in C, 2^lambda=lambda^++$.
As far as I know, if we require $theta=cf(kappa)> omega_1$, the large cardinal assumption is optimal, but for $theta=omega_1,$ I think it is open if this assumption is optimal.
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
$endgroup$
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
1
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
$endgroup$
– Mohammad Golshani
Apr 16 at 4:18
1
$begingroup$
Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
$endgroup$
– Mohammad Golshani
Apr 16 at 4:20
|
show 8 more comments
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$begingroup$
Suppose $kappa$ is a singular cardinal and there are $cf(kappa)$-many measurable cardinals $lambda < kappa$ with $o(lambda)=lambda^++$ cofinal in $kappa.$ Then you can perform a Prikry type iteration and get the failure of $SCH$ at cofinally many singular cardinals below $kappa.$
Now suppose we also want for $SCH$ to fail at $kappa$ itself. First let us consider the countable cofinality.
Assume $kappa$ is a measurable cardinal with $o(kappa)=kappa^+++1.$ Then we can get an extension in which $cf(kappa)=omega, 2^kappa=kappa^++$ and for cofinally many singular cardinals $lambda$ below $kappa$, we have $2^lambda=lambda^++$. I don't know if this assumption is really needed or if it can be reduced.
For uncountable cofinality, say $theta$, a measurable cardinal $kappa$ with $o(kappa)=kappa^+++theta$ is sufficient. As then you can first find an extension in which $2^kappa=kappa^++$ and such that in the extension, $o(kappa)=theta$. Then if you perform Magidor forcing for changing cofinality of $kappa$ to $theta,$ you can get a club $C$ of singular cardinals below $kappa$ such that for all $lambda in C, 2^lambda=lambda^++$.
As far as I know, if we require $theta=cf(kappa)> omega_1$, the large cardinal assumption is optimal, but for $theta=omega_1,$ I think it is open if this assumption is optimal.
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
$endgroup$
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
1
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
$endgroup$
– Mohammad Golshani
Apr 16 at 4:18
1
$begingroup$
Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
$endgroup$
– Mohammad Golshani
Apr 16 at 4:20
|
show 8 more comments
$begingroup$
Suppose $kappa$ is a singular cardinal and there are $cf(kappa)$-many measurable cardinals $lambda < kappa$ with $o(lambda)=lambda^++$ cofinal in $kappa.$ Then you can perform a Prikry type iteration and get the failure of $SCH$ at cofinally many singular cardinals below $kappa.$
Now suppose we also want for $SCH$ to fail at $kappa$ itself. First let us consider the countable cofinality.
Assume $kappa$ is a measurable cardinal with $o(kappa)=kappa^+++1.$ Then we can get an extension in which $cf(kappa)=omega, 2^kappa=kappa^++$ and for cofinally many singular cardinals $lambda$ below $kappa$, we have $2^lambda=lambda^++$. I don't know if this assumption is really needed or if it can be reduced.
For uncountable cofinality, say $theta$, a measurable cardinal $kappa$ with $o(kappa)=kappa^+++theta$ is sufficient. As then you can first find an extension in which $2^kappa=kappa^++$ and such that in the extension, $o(kappa)=theta$. Then if you perform Magidor forcing for changing cofinality of $kappa$ to $theta,$ you can get a club $C$ of singular cardinals below $kappa$ such that for all $lambda in C, 2^lambda=lambda^++$.
As far as I know, if we require $theta=cf(kappa)> omega_1$, the large cardinal assumption is optimal, but for $theta=omega_1,$ I think it is open if this assumption is optimal.
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
$endgroup$
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
1
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
$endgroup$
– Mohammad Golshani
Apr 16 at 4:18
1
$begingroup$
Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
$endgroup$
– Mohammad Golshani
Apr 16 at 4:20
|
show 8 more comments
$begingroup$
Suppose $kappa$ is a singular cardinal and there are $cf(kappa)$-many measurable cardinals $lambda < kappa$ with $o(lambda)=lambda^++$ cofinal in $kappa.$ Then you can perform a Prikry type iteration and get the failure of $SCH$ at cofinally many singular cardinals below $kappa.$
Now suppose we also want for $SCH$ to fail at $kappa$ itself. First let us consider the countable cofinality.
Assume $kappa$ is a measurable cardinal with $o(kappa)=kappa^+++1.$ Then we can get an extension in which $cf(kappa)=omega, 2^kappa=kappa^++$ and for cofinally many singular cardinals $lambda$ below $kappa$, we have $2^lambda=lambda^++$. I don't know if this assumption is really needed or if it can be reduced.
For uncountable cofinality, say $theta$, a measurable cardinal $kappa$ with $o(kappa)=kappa^+++theta$ is sufficient. As then you can first find an extension in which $2^kappa=kappa^++$ and such that in the extension, $o(kappa)=theta$. Then if you perform Magidor forcing for changing cofinality of $kappa$ to $theta,$ you can get a club $C$ of singular cardinals below $kappa$ such that for all $lambda in C, 2^lambda=lambda^++$.
As far as I know, if we require $theta=cf(kappa)> omega_1$, the large cardinal assumption is optimal, but for $theta=omega_1,$ I think it is open if this assumption is optimal.
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
$endgroup$
Suppose $kappa$ is a singular cardinal and there are $cf(kappa)$-many measurable cardinals $lambda < kappa$ with $o(lambda)=lambda^++$ cofinal in $kappa.$ Then you can perform a Prikry type iteration and get the failure of $SCH$ at cofinally many singular cardinals below $kappa.$
Now suppose we also want for $SCH$ to fail at $kappa$ itself. First let us consider the countable cofinality.
Assume $kappa$ is a measurable cardinal with $o(kappa)=kappa^+++1.$ Then we can get an extension in which $cf(kappa)=omega, 2^kappa=kappa^++$ and for cofinally many singular cardinals $lambda$ below $kappa$, we have $2^lambda=lambda^++$. I don't know if this assumption is really needed or if it can be reduced.
For uncountable cofinality, say $theta$, a measurable cardinal $kappa$ with $o(kappa)=kappa^+++theta$ is sufficient. As then you can first find an extension in which $2^kappa=kappa^++$ and such that in the extension, $o(kappa)=theta$. Then if you perform Magidor forcing for changing cofinality of $kappa$ to $theta,$ you can get a club $C$ of singular cardinals below $kappa$ such that for all $lambda in C, 2^lambda=lambda^++$.
As far as I know, if we require $theta=cf(kappa)> omega_1$, the large cardinal assumption is optimal, but for $theta=omega_1,$ I think it is open if this assumption is optimal.
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
edited Apr 15 at 10:31
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
answered Apr 15 at 10:09
Mohammad GolshaniMohammad Golshani
20.1k2 gold badges68 silver badges155 bronze badges
20.1k2 gold badges68 silver badges155 bronze badges
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
1
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
$endgroup$
– Mohammad Golshani
Apr 16 at 4:18
1
$begingroup$
Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
$endgroup$
– Mohammad Golshani
Apr 16 at 4:20
|
show 8 more comments
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
1
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
$endgroup$
– Mohammad Golshani
Apr 16 at 4:18
1
$begingroup$
Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
$endgroup$
– Mohammad Golshani
Apr 16 at 4:20
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Note that you may assume that the sequence does not contain its limit points.
$endgroup$
– Mohammad Golshani
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
Okay. I sort of expected that to happen. What about the actual question?
$endgroup$
– Asaf Karagila
Apr 15 at 10:16
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
$begingroup$
What is the question?
$endgroup$
– Mohammad Golshani
Apr 15 at 10:17
1
1
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
$endgroup$
– Mohammad Golshani
Apr 16 at 4:18
$begingroup$
I assume you say something like this: If for example $o(kappa)=kappa^+++omega_3+3$, then first we can find a model with $2^kappa=kappa^+++o(kappa)=omega_3+3$. Then we can make $cf(kappa)=omega$ by a cofinal sequence $(kappa_n: n<omega)$ of point with $o(kappa_n)=omega_3+2.$ then we can add an $omega$-sequence cofinal in each of these $kappa_n$'s, say $(kappa_n, m: m<omega)$ consisting of points of $o(kappa_n, m)=omega_3+1.$
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– Mohammad Golshani
Apr 16 at 4:18
1
1
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Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
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– Mohammad Golshani
Apr 16 at 4:20
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Repeat once more, and find the sequence $(kappa_n,m,l: l<omega)$, cofinal in $kappa_n, m$ with $o(kappa_n, m, l)=omega_3.$ Now apply Magidor's forcing to change the cofinality of each $kappa_n, m, l$ to $omega_3$.
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– Mohammad Golshani
Apr 16 at 4:20
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Could you remind the set-theorist dilettantes what $o(kappa)$ stands for?
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– Gro-Tsen
Apr 15 at 10:27
3
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@Gro-Tsen: It's the Mitchell order. In some sense, it tells you how large is a measurable. $o(kappa)=1$ means just being a measurable, but $o(kappa)=2$ means that there is a normal measure which concentrates on those $o(lambda)=1$. Which in turn means that the ultrapower by that measure preserves the measurability of $kappa$ itself. You can read more here: cantorsattic.info/Mitchell_rank
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– Asaf Karagila
Apr 15 at 11:11