Mixing
Can elementary embeddings make ordinals jump over strongly inaccessible cardinals?
$begingroup$
Let $j:Vto M$ be an elementary embedding. Assume $xigeq crit(j)$ is strongly inaccessible, and $alpha<xi$. Is it possible that $j(alpha)geqxi$ ?
If yes - does the additional assumption $V_xi subset M$ change it?
Note that if $j$ is an ultrapower embedding ($j=j_U$ for some ultrafilter $U$), then the answer is indeed "no" - if $xi>crit(j)$ is strongly inaccessible then $j(xi)=xi$ so $alpha<xi$ implies $j(alpha)<xi$.
set-theory large-cardinals
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
$endgroup$
add a comment |
$begingroup$
Let $j:Vto M$ be an elementary embedding. Assume $xigeq crit(j)$ is strongly inaccessible, and $alpha<xi$. Is it possible that $j(alpha)geqxi$ ?
If yes - does the additional assumption $V_xi subset M$ change it?
Note that if $j$ is an ultrapower embedding ($j=j_U$ for some ultrafilter $U$), then the answer is indeed "no" - if $xi>crit(j)$ is strongly inaccessible then $j(xi)=xi$ so $alpha<xi$ implies $j(alpha)<xi$.
set-theory large-cardinals
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
$endgroup$
add a comment |
$begingroup$
Let $j:Vto M$ be an elementary embedding. Assume $xigeq crit(j)$ is strongly inaccessible, and $alpha<xi$. Is it possible that $j(alpha)geqxi$ ?
If yes - does the additional assumption $V_xi subset M$ change it?
Note that if $j$ is an ultrapower embedding ($j=j_U$ for some ultrafilter $U$), then the answer is indeed "no" - if $xi>crit(j)$ is strongly inaccessible then $j(xi)=xi$ so $alpha<xi$ implies $j(alpha)<xi$.
set-theory large-cardinals
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
$endgroup$
Let $j:Vto M$ be an elementary embedding. Assume $xigeq crit(j)$ is strongly inaccessible, and $alpha<xi$. Is it possible that $j(alpha)geqxi$ ?
If yes - does the additional assumption $V_xi subset M$ change it?
Note that if $j$ is an ultrapower embedding ($j=j_U$ for some ultrafilter $U$), then the answer is indeed "no" - if $xi>crit(j)$ is strongly inaccessible then $j(xi)=xi$ so $alpha<xi$ implies $j(alpha)<xi$.
set-theory large-cardinals
set-theory large-cardinals
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
edited Jan 16 at 13:07
Asaf Karagila♦
305k33435765
305k33435765
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
asked Jan 16 at 13:00
Ur Ya'arUr Ya'ar
544211
544211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Of course it's possible, with sufficiently large large cardinals.
If $kappa$ is supercompact and $xi>kappa$ is inaccessible, then there is an elementary embedding $j$ such that $j(kappa)>xi$, in fact, there is one with $kappa$ as the critical point. In fact, in that scenario, we can ask the target model to be closed under $xi$-sequences, which in case $xi$ is a strong limit, we get $V_xisubseteq M$.
Superstrong cardinals are those where $V_{j(kappa)}subseteq M$, and those are weaker than supercompact.
And finally, note that while it's true that supercompact embeddings are not ultrapowers like measurable cardinals are, but they are ultrapowers by larger sets and "better" measures: namely $mathcal P_kappa(xi)$ and some fine and normal $kappa$-complete measure on it.
In fact, just asking that $j(kappa)$ is arbitrary large can be done with just a single measurable cardinal. We simply pick an embedding (your favorite ultrapower, for example), and iterate it. Gaifman's theorem tells us that the direct limits are well-founded as well, and it is easy to show that the critical points must always strictly increase.
This is also a way of building reasonably canonical inner models. If you start with the smallest measurable cardinal in your universe, and you keep iterating the embedding, then you necessarily kill your measurable cardinals, one at a time. So the result is an inner model which does not have a measurable cardinal in it. And in fact, this is a way of building the Dodd-Jensen core model.
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
$endgroup$
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
1
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
2
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
|
show 1 more comment
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1 Answer
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$begingroup$
Of course it's possible, with sufficiently large large cardinals.
If $kappa$ is supercompact and $xi>kappa$ is inaccessible, then there is an elementary embedding $j$ such that $j(kappa)>xi$, in fact, there is one with $kappa$ as the critical point. In fact, in that scenario, we can ask the target model to be closed under $xi$-sequences, which in case $xi$ is a strong limit, we get $V_xisubseteq M$.
Superstrong cardinals are those where $V_{j(kappa)}subseteq M$, and those are weaker than supercompact.
And finally, note that while it's true that supercompact embeddings are not ultrapowers like measurable cardinals are, but they are ultrapowers by larger sets and "better" measures: namely $mathcal P_kappa(xi)$ and some fine and normal $kappa$-complete measure on it.
In fact, just asking that $j(kappa)$ is arbitrary large can be done with just a single measurable cardinal. We simply pick an embedding (your favorite ultrapower, for example), and iterate it. Gaifman's theorem tells us that the direct limits are well-founded as well, and it is easy to show that the critical points must always strictly increase.
This is also a way of building reasonably canonical inner models. If you start with the smallest measurable cardinal in your universe, and you keep iterating the embedding, then you necessarily kill your measurable cardinals, one at a time. So the result is an inner model which does not have a measurable cardinal in it. And in fact, this is a way of building the Dodd-Jensen core model.
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
$endgroup$
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
1
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
2
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
|
show 1 more comment
$begingroup$
Of course it's possible, with sufficiently large large cardinals.
If $kappa$ is supercompact and $xi>kappa$ is inaccessible, then there is an elementary embedding $j$ such that $j(kappa)>xi$, in fact, there is one with $kappa$ as the critical point. In fact, in that scenario, we can ask the target model to be closed under $xi$-sequences, which in case $xi$ is a strong limit, we get $V_xisubseteq M$.
Superstrong cardinals are those where $V_{j(kappa)}subseteq M$, and those are weaker than supercompact.
And finally, note that while it's true that supercompact embeddings are not ultrapowers like measurable cardinals are, but they are ultrapowers by larger sets and "better" measures: namely $mathcal P_kappa(xi)$ and some fine and normal $kappa$-complete measure on it.
In fact, just asking that $j(kappa)$ is arbitrary large can be done with just a single measurable cardinal. We simply pick an embedding (your favorite ultrapower, for example), and iterate it. Gaifman's theorem tells us that the direct limits are well-founded as well, and it is easy to show that the critical points must always strictly increase.
This is also a way of building reasonably canonical inner models. If you start with the smallest measurable cardinal in your universe, and you keep iterating the embedding, then you necessarily kill your measurable cardinals, one at a time. So the result is an inner model which does not have a measurable cardinal in it. And in fact, this is a way of building the Dodd-Jensen core model.
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
$endgroup$
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
1
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
2
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
|
show 1 more comment
$begingroup$
Of course it's possible, with sufficiently large large cardinals.
If $kappa$ is supercompact and $xi>kappa$ is inaccessible, then there is an elementary embedding $j$ such that $j(kappa)>xi$, in fact, there is one with $kappa$ as the critical point. In fact, in that scenario, we can ask the target model to be closed under $xi$-sequences, which in case $xi$ is a strong limit, we get $V_xisubseteq M$.
Superstrong cardinals are those where $V_{j(kappa)}subseteq M$, and those are weaker than supercompact.
And finally, note that while it's true that supercompact embeddings are not ultrapowers like measurable cardinals are, but they are ultrapowers by larger sets and "better" measures: namely $mathcal P_kappa(xi)$ and some fine and normal $kappa$-complete measure on it.
In fact, just asking that $j(kappa)$ is arbitrary large can be done with just a single measurable cardinal. We simply pick an embedding (your favorite ultrapower, for example), and iterate it. Gaifman's theorem tells us that the direct limits are well-founded as well, and it is easy to show that the critical points must always strictly increase.
This is also a way of building reasonably canonical inner models. If you start with the smallest measurable cardinal in your universe, and you keep iterating the embedding, then you necessarily kill your measurable cardinals, one at a time. So the result is an inner model which does not have a measurable cardinal in it. And in fact, this is a way of building the Dodd-Jensen core model.
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
$endgroup$
Of course it's possible, with sufficiently large large cardinals.
If $kappa$ is supercompact and $xi>kappa$ is inaccessible, then there is an elementary embedding $j$ such that $j(kappa)>xi$, in fact, there is one with $kappa$ as the critical point. In fact, in that scenario, we can ask the target model to be closed under $xi$-sequences, which in case $xi$ is a strong limit, we get $V_xisubseteq M$.
Superstrong cardinals are those where $V_{j(kappa)}subseteq M$, and those are weaker than supercompact.
And finally, note that while it's true that supercompact embeddings are not ultrapowers like measurable cardinals are, but they are ultrapowers by larger sets and "better" measures: namely $mathcal P_kappa(xi)$ and some fine and normal $kappa$-complete measure on it.
In fact, just asking that $j(kappa)$ is arbitrary large can be done with just a single measurable cardinal. We simply pick an embedding (your favorite ultrapower, for example), and iterate it. Gaifman's theorem tells us that the direct limits are well-founded as well, and it is easy to show that the critical points must always strictly increase.
This is also a way of building reasonably canonical inner models. If you start with the smallest measurable cardinal in your universe, and you keep iterating the embedding, then you necessarily kill your measurable cardinals, one at a time. So the result is an inner model which does not have a measurable cardinal in it. And in fact, this is a way of building the Dodd-Jensen core model.
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
edited Jan 16 at 15:21
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
answered Jan 16 at 13:07
Asaf Karagila♦Asaf Karagila
305k33435765
305k33435765
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
1
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
2
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
|
show 1 more comment
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
1
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
2
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
oh yeh, I probably should have thought in that direction.
$endgroup$
– Ur Ya'ar
Jan 16 at 13:13
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
$begingroup$
So, what's the consistency strength of such an embedding? say to jump over the first inaccessible over the critical point?
$endgroup$
– Ur Ya'ar
Jan 16 at 13:15
1
1
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
Well. Just an embedding is actually a measurable. You can iterate the embedding until it goes above the inaccessible. If you want to capture some part of the universe, then some strongness or superstrongnesss are necessary.
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:20
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
$begingroup$
(I should probably edit that in once I'm back in my office.)
$endgroup$
– Asaf Karagila♦
Jan 16 at 13:23
2
2
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
$begingroup$
Yes. In fact, this is one of the way of building a core model. You start with the smallest measurable you have, and you iterate the embedding through the ordinals. The end is the Dodd-Jensen core model.
$endgroup$
– Asaf Karagila♦
Jan 16 at 15:16
|
show 1 more comment
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