amenable ultrafilters
$begingroup$
Suppose $M$ is a transitive model of ZFC-powerset. If $kappa in M$ is a cardinal and $U$ is an ultrafilter on the boolean algebra $mathcal P(kappa)^M$, we say $U$ is amenable to $M$ if whenever ${ X_alpha : alpha < kappa } subseteq mathcal P(kappa)$ is in $M$, then ${ alpha : X_alpha in U } in M$. In these notes, Steel says that if $j : M to N$ is elementary with critical point $kappa$, and $U$ is the ultrafilter on $mathcal P(kappa)^M$ derived from $j$, then $U$ is amenable to $M$ if and only if $mathcal P(kappa)^M = mathcal P(kappa)^N$. I am able to show that $U$ is amenable to $M$ iff $mathcal P(kappa)^M = mathcal P(kappa)^{Ult(M,U)}$, but I don't see why necessarily $mathcal P(kappa)^{Ult(M,U)} = P(kappa)^N$. Is the claim true, and how do you show it?
logic set-theory large-cardinals
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
$endgroup$
add a comment |
$begingroup$
Suppose $M$ is a transitive model of ZFC-powerset. If $kappa in M$ is a cardinal and $U$ is an ultrafilter on the boolean algebra $mathcal P(kappa)^M$, we say $U$ is amenable to $M$ if whenever ${ X_alpha : alpha < kappa } subseteq mathcal P(kappa)$ is in $M$, then ${ alpha : X_alpha in U } in M$. In these notes, Steel says that if $j : M to N$ is elementary with critical point $kappa$, and $U$ is the ultrafilter on $mathcal P(kappa)^M$ derived from $j$, then $U$ is amenable to $M$ if and only if $mathcal P(kappa)^M = mathcal P(kappa)^N$. I am able to show that $U$ is amenable to $M$ iff $mathcal P(kappa)^M = mathcal P(kappa)^{Ult(M,U)}$, but I don't see why necessarily $mathcal P(kappa)^{Ult(M,U)} = P(kappa)^N$. Is the claim true, and how do you show it?
logic set-theory large-cardinals
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
$endgroup$
$begingroup$
Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $kappa$ itself.
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 6:09
$begingroup$
@AsafKaragila I need to show the factor map critical point is above $2^kappa$. Why is that?
$endgroup$
– mbsq
Dec 15 '18 at 6:58
add a comment |
$begingroup$
Suppose $M$ is a transitive model of ZFC-powerset. If $kappa in M$ is a cardinal and $U$ is an ultrafilter on the boolean algebra $mathcal P(kappa)^M$, we say $U$ is amenable to $M$ if whenever ${ X_alpha : alpha < kappa } subseteq mathcal P(kappa)$ is in $M$, then ${ alpha : X_alpha in U } in M$. In these notes, Steel says that if $j : M to N$ is elementary with critical point $kappa$, and $U$ is the ultrafilter on $mathcal P(kappa)^M$ derived from $j$, then $U$ is amenable to $M$ if and only if $mathcal P(kappa)^M = mathcal P(kappa)^N$. I am able to show that $U$ is amenable to $M$ iff $mathcal P(kappa)^M = mathcal P(kappa)^{Ult(M,U)}$, but I don't see why necessarily $mathcal P(kappa)^{Ult(M,U)} = P(kappa)^N$. Is the claim true, and how do you show it?
logic set-theory large-cardinals
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
$endgroup$
Suppose $M$ is a transitive model of ZFC-powerset. If $kappa in M$ is a cardinal and $U$ is an ultrafilter on the boolean algebra $mathcal P(kappa)^M$, we say $U$ is amenable to $M$ if whenever ${ X_alpha : alpha < kappa } subseteq mathcal P(kappa)$ is in $M$, then ${ alpha : X_alpha in U } in M$. In these notes, Steel says that if $j : M to N$ is elementary with critical point $kappa$, and $U$ is the ultrafilter on $mathcal P(kappa)^M$ derived from $j$, then $U$ is amenable to $M$ if and only if $mathcal P(kappa)^M = mathcal P(kappa)^N$. I am able to show that $U$ is amenable to $M$ iff $mathcal P(kappa)^M = mathcal P(kappa)^{Ult(M,U)}$, but I don't see why necessarily $mathcal P(kappa)^{Ult(M,U)} = P(kappa)^N$. Is the claim true, and how do you show it?
logic set-theory large-cardinals
logic set-theory large-cardinals
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
asked Dec 14 '18 at 16:22
mbsqmbsq
1,054512
1,054512
$begingroup$
Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $kappa$ itself.
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 6:09
$begingroup$
@AsafKaragila I need to show the factor map critical point is above $2^kappa$. Why is that?
$endgroup$
– mbsq
Dec 15 '18 at 6:58
add a comment |
$begingroup$
Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $kappa$ itself.
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 6:09
$begingroup$
@AsafKaragila I need to show the factor map critical point is above $2^kappa$. Why is that?
$endgroup$
– mbsq
Dec 15 '18 at 6:58
$begingroup$
Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $kappa$ itself.
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 6:09
$begingroup$
Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $kappa$ itself.
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 6:09
$begingroup$
@AsafKaragila I need to show the factor map critical point is above $2^kappa$. Why is that?
$endgroup$
– mbsq
Dec 15 '18 at 6:58
$begingroup$
@AsafKaragila I need to show the factor map critical point is above $2^kappa$. Why is that?
$endgroup$
– mbsq
Dec 15 '18 at 6:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The claim is false.
Assume there is a measurable $kappa$ and for every inaccessible $delta < kappa$, there is a precipitous ideal on $delta^+$. A model of this can be obtained by forcing from a model with a supercompact and a measurable above it.
Let $i : V to M$ be the ultrapower embedding by a normal measure $U$ on $kappa$. In $M$, there is a precipitous ideal $I$ on $kappa^+$. Let $k : M to N$ be a generic embedding coming from $I$. Note that $crit(k) = kappa^+$. Let $j = k circ i$.
$U$ is amenable to $V$, and for every $A in mathcal P(kappa)^V$, $kappa in i(A)$ iff $kappa in j(A)$. So $U$ is also the ultrafilter derived from $j$. But $mathcal P(kappa)^M not= mathcal P(kappa)^N$, since forcing with $I$ collapses $kappa^+$.
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
$endgroup$
1
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
The claim is false.
Assume there is a measurable $kappa$ and for every inaccessible $delta < kappa$, there is a precipitous ideal on $delta^+$. A model of this can be obtained by forcing from a model with a supercompact and a measurable above it.
Let $i : V to M$ be the ultrapower embedding by a normal measure $U$ on $kappa$. In $M$, there is a precipitous ideal $I$ on $kappa^+$. Let $k : M to N$ be a generic embedding coming from $I$. Note that $crit(k) = kappa^+$. Let $j = k circ i$.
$U$ is amenable to $V$, and for every $A in mathcal P(kappa)^V$, $kappa in i(A)$ iff $kappa in j(A)$. So $U$ is also the ultrafilter derived from $j$. But $mathcal P(kappa)^M not= mathcal P(kappa)^N$, since forcing with $I$ collapses $kappa^+$.
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
$endgroup$
1
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
add a comment |
$begingroup$
The claim is false.
Assume there is a measurable $kappa$ and for every inaccessible $delta < kappa$, there is a precipitous ideal on $delta^+$. A model of this can be obtained by forcing from a model with a supercompact and a measurable above it.
Let $i : V to M$ be the ultrapower embedding by a normal measure $U$ on $kappa$. In $M$, there is a precipitous ideal $I$ on $kappa^+$. Let $k : M to N$ be a generic embedding coming from $I$. Note that $crit(k) = kappa^+$. Let $j = k circ i$.
$U$ is amenable to $V$, and for every $A in mathcal P(kappa)^V$, $kappa in i(A)$ iff $kappa in j(A)$. So $U$ is also the ultrafilter derived from $j$. But $mathcal P(kappa)^M not= mathcal P(kappa)^N$, since forcing with $I$ collapses $kappa^+$.
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
$endgroup$
1
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
add a comment |
$begingroup$
The claim is false.
Assume there is a measurable $kappa$ and for every inaccessible $delta < kappa$, there is a precipitous ideal on $delta^+$. A model of this can be obtained by forcing from a model with a supercompact and a measurable above it.
Let $i : V to M$ be the ultrapower embedding by a normal measure $U$ on $kappa$. In $M$, there is a precipitous ideal $I$ on $kappa^+$. Let $k : M to N$ be a generic embedding coming from $I$. Note that $crit(k) = kappa^+$. Let $j = k circ i$.
$U$ is amenable to $V$, and for every $A in mathcal P(kappa)^V$, $kappa in i(A)$ iff $kappa in j(A)$. So $U$ is also the ultrafilter derived from $j$. But $mathcal P(kappa)^M not= mathcal P(kappa)^N$, since forcing with $I$ collapses $kappa^+$.
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
$endgroup$
The claim is false.
Assume there is a measurable $kappa$ and for every inaccessible $delta < kappa$, there is a precipitous ideal on $delta^+$. A model of this can be obtained by forcing from a model with a supercompact and a measurable above it.
Let $i : V to M$ be the ultrapower embedding by a normal measure $U$ on $kappa$. In $M$, there is a precipitous ideal $I$ on $kappa^+$. Let $k : M to N$ be a generic embedding coming from $I$. Note that $crit(k) = kappa^+$. Let $j = k circ i$.
$U$ is amenable to $V$, and for every $A in mathcal P(kappa)^V$, $kappa in i(A)$ iff $kappa in j(A)$. So $U$ is also the ultrafilter derived from $j$. But $mathcal P(kappa)^M not= mathcal P(kappa)^N$, since forcing with $I$ collapses $kappa^+$.
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
answered Dec 15 '18 at 10:21
mbsqmbsq
1,054512
1,054512
1
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
add a comment |
1
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
1
1
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Very nice! I can finally move on with my day...
$endgroup$
– Miha Habič
Dec 15 '18 at 10:27
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
Do you really need a supercompact? Wouldn't a measurable with $o(kappa)=1$ (or $2$ if $1$ means just measurable for your count) be enough? Just iterate with Easton support Levy collapses of a large set of measurable cardinals to be successor of inaccessible cardinals and then you get the ideals to exist by the usual chain condition argument?
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 13:51
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
@AsafKaragila I want to preserve measurability so the obvious thing to do is collapse the first measurable above kappa in the last step. It’s probably overkill.
$endgroup$
– mbsq
Dec 15 '18 at 13:55
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
$begingroup$
Maybe Woodin is enough using “surgery”?
$endgroup$
– mbsq
Dec 15 '18 at 14:25
add a comment |
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$begingroup$
Factor $j$ into an ultrapower and a second embedding, and show the second embedding has a critical point above $kappa$ itself.
$endgroup$
– Asaf Karagila♦
Dec 15 '18 at 6:09
$begingroup$
@AsafKaragila I need to show the factor map critical point is above $2^kappa$. Why is that?
$endgroup$
– mbsq
Dec 15 '18 at 6:58