Mixing
The transitive collapse of an ultraproduct of $V$ is an inner model
$begingroup$
So this is the situation: Let $U$ be an ultrafilter over a set $S$ and let $M_U$ be the transitive collapse of $text{Ult}(V, U)$. $M_U$ is an inner model of ZFC.
Thus far i have used Los's theorem to prove that for every sentence $varphi$, $text{Ult}(V, U) models varphi leftrightarrow {i in S: V models varphi } in U$, which was easy and shows that $M_U$ is a model of ZFC. But now i have no idea on how to show that $text{Ord} subseteq M_U$. I would be really grateful for any hints or answers.
set-theory model-theory large-cardinals
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
$endgroup$
add a comment |
$begingroup$
So this is the situation: Let $U$ be an ultrafilter over a set $S$ and let $M_U$ be the transitive collapse of $text{Ult}(V, U)$. $M_U$ is an inner model of ZFC.
Thus far i have used Los's theorem to prove that for every sentence $varphi$, $text{Ult}(V, U) models varphi leftrightarrow {i in S: V models varphi } in U$, which was easy and shows that $M_U$ is a model of ZFC. But now i have no idea on how to show that $text{Ord} subseteq M_U$. I would be really grateful for any hints or answers.
set-theory model-theory large-cardinals
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
$endgroup$
add a comment |
$begingroup$
So this is the situation: Let $U$ be an ultrafilter over a set $S$ and let $M_U$ be the transitive collapse of $text{Ult}(V, U)$. $M_U$ is an inner model of ZFC.
Thus far i have used Los's theorem to prove that for every sentence $varphi$, $text{Ult}(V, U) models varphi leftrightarrow {i in S: V models varphi } in U$, which was easy and shows that $M_U$ is a model of ZFC. But now i have no idea on how to show that $text{Ord} subseteq M_U$. I would be really grateful for any hints or answers.
set-theory model-theory large-cardinals
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
$endgroup$
So this is the situation: Let $U$ be an ultrafilter over a set $S$ and let $M_U$ be the transitive collapse of $text{Ult}(V, U)$. $M_U$ is an inner model of ZFC.
Thus far i have used Los's theorem to prove that for every sentence $varphi$, $text{Ult}(V, U) models varphi leftrightarrow {i in S: V models varphi } in U$, which was easy and shows that $M_U$ is a model of ZFC. But now i have no idea on how to show that $text{Ord} subseteq M_U$. I would be really grateful for any hints or answers.
set-theory model-theory large-cardinals
set-theory model-theory large-cardinals
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
edited Jan 24 at 16:29
Asaf Karagila♦
306k33438769
306k33438769
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
asked Jan 24 at 11:38
Shervin SorouriShervin Sorouri
519211
519211
add a comment |
add a comment |
1 Answer
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$begingroup$
Either show that the embedding from $V$ to $M_U$ does not lower ranks.
Or else simply note that $M_U$ is unbounded in size (as the embedding is injective) and for any $x in M_U$, $operatorname{rk}(x) in M_U$ since $M_U$ satisfies ZFC (and $operatorname{rk}$ is absolute for transitive models).
answered Jan 24 at 12:03
JonathanJonathan
57948
$endgroup$
2
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Either show that the embedding from $V$ to $M_U$ does not lower ranks.
Or else simply note that $M_U$ is unbounded in size (as the embedding is injective) and for any $x in M_U$, $operatorname{rk}(x) in M_U$ since $M_U$ satisfies ZFC (and $operatorname{rk}$ is absolute for transitive models).
answered Jan 24 at 12:03
JonathanJonathan
57948
$endgroup$
2
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
add a comment |
$begingroup$
Either show that the embedding from $V$ to $M_U$ does not lower ranks.
Or else simply note that $M_U$ is unbounded in size (as the embedding is injective) and for any $x in M_U$, $operatorname{rk}(x) in M_U$ since $M_U$ satisfies ZFC (and $operatorname{rk}$ is absolute for transitive models).
answered Jan 24 at 12:03
JonathanJonathan
57948
$endgroup$
2
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
add a comment |
$begingroup$
Either show that the embedding from $V$ to $M_U$ does not lower ranks.
Or else simply note that $M_U$ is unbounded in size (as the embedding is injective) and for any $x in M_U$, $operatorname{rk}(x) in M_U$ since $M_U$ satisfies ZFC (and $operatorname{rk}$ is absolute for transitive models).
answered Jan 24 at 12:03
JonathanJonathan
57948
$endgroup$
Either show that the embedding from $V$ to $M_U$ does not lower ranks.
Or else simply note that $M_U$ is unbounded in size (as the embedding is injective) and for any $x in M_U$, $operatorname{rk}(x) in M_U$ since $M_U$ satisfies ZFC (and $operatorname{rk}$ is absolute for transitive models).
answered Jan 24 at 12:03
JonathanJonathan
57948
answered Jan 24 at 12:03
JonathanJonathan
57948
answered Jan 24 at 12:03
JonathanJonathan
57948
answered Jan 24 at 12:03
JonathanJonathan
57948
57948
2
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
add a comment |
2
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
2
2
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
$begingroup$
I think it is easier to see that the ordinals of the ultraproduct collapse to actual ordinals, and there is a proper class of them, as can be seen by considering constant functions.
$endgroup$
– Andrés E. Caicedo
Jan 25 at 1:46
add a comment |
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