Existence of a transitive model is strictly stronger than consistency?

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It seems like we should be able to prove that the existence of a transitive model for ZFC is strictly stronger than Con(ZFC), but I can't find anything saying so / giving an argument for it. Is there a standard way of demonstrating this?

An example of what I'm looking for, if it existed: given Con(ZFC), is there a way of generating a model which models Con(ZFC) but which believes that no models of ZFC are transitive?

logic reference-request set-theory

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asked Aug 19 at 1:24

Spencer

1998

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up vote
6
down vote

favorite

It seems like we should be able to prove that the existence of a transitive model for ZFC is strictly stronger than Con(ZFC), but I can't find anything saying so / giving an argument for it. Is there a standard way of demonstrating this?

An example of what I'm looking for, if it existed: given Con(ZFC), is there a way of generating a model which models Con(ZFC) but which believes that no models of ZFC are transitive?

logic reference-request set-theory

share|cite|improve this question

asked Aug 19 at 1:24

Spencer

1998

add a comment | 

up vote
6
down vote

favorite

up vote
6
down vote

favorite

It seems like we should be able to prove that the existence of a transitive model for ZFC is strictly stronger than Con(ZFC), but I can't find anything saying so / giving an argument for it. Is there a standard way of demonstrating this?

An example of what I'm looking for, if it existed: given Con(ZFC), is there a way of generating a model which models Con(ZFC) but which believes that no models of ZFC are transitive?

logic reference-request set-theory

share|cite|improve this question

asked Aug 19 at 1:24

Spencer

1998

It seems like we should be able to prove that the existence of a transitive model for ZFC is strictly stronger than Con(ZFC), but I can't find anything saying so / giving an argument for it. Is there a standard way of demonstrating this?

An example of what I'm looking for, if it existed: given Con(ZFC), is there a way of generating a model which models Con(ZFC) but which believes that no models of ZFC are transitive?

logic reference-request set-theory

share|cite|improve this question

asked Aug 19 at 1:24

Spencer

1998

share|cite|improve this question

share|cite|improve this question

asked Aug 19 at 1:24

Spencer

1998

asked Aug 19 at 1:24

Spencer

1998

asked Aug 19 at 1:24

Spencer

1998

1998

add a comment | 

add a comment | 

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Any model of ZFC+Con(ZFC)+$neg$Con(ZFC+Con(ZFC)) should do. (This theory is, by the second incompleteness theorem, consistent if ZFC+Con(ZFC) is).

Because the model believes Con(ZFC) and also believes that no model of ZFC can believe Con(ZFC), the only models of ZFC it can know will be ones that it considers to have non-standard integers. Non-standard models of arithmetic are never well-founded, so such an internal model cannot be transitive.

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edited Aug 19 at 2:01

answered Aug 19 at 1:46

Henning Makholm

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1 Answer
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1 Answer
1

active

oldest

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active

oldest

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active

oldest

votes

up vote
10
down vote



accepted

Any model of ZFC+Con(ZFC)+$neg$Con(ZFC+Con(ZFC)) should do. (This theory is, by the second incompleteness theorem, consistent if ZFC+Con(ZFC) is).

Because the model believes Con(ZFC) and also believes that no model of ZFC can believe Con(ZFC), the only models of ZFC it can know will be ones that it considers to have non-standard integers. Non-standard models of arithmetic are never well-founded, so such an internal model cannot be transitive.

share|cite|improve this answer

edited Aug 19 at 2:01

answered Aug 19 at 1:46

Henning Makholm

229k16295526

add a comment | 

up vote
10
down vote



accepted

Any model of ZFC+Con(ZFC)+$neg$Con(ZFC+Con(ZFC)) should do. (This theory is, by the second incompleteness theorem, consistent if ZFC+Con(ZFC) is).

Because the model believes Con(ZFC) and also believes that no model of ZFC can believe Con(ZFC), the only models of ZFC it can know will be ones that it considers to have non-standard integers. Non-standard models of arithmetic are never well-founded, so such an internal model cannot be transitive.

share|cite|improve this answer

edited Aug 19 at 2:01

answered Aug 19 at 1:46

Henning Makholm

229k16295526

add a comment | 

up vote
10
down vote



accepted

up vote
10
down vote



accepted

Any model of ZFC+Con(ZFC)+$neg$Con(ZFC+Con(ZFC)) should do. (This theory is, by the second incompleteness theorem, consistent if ZFC+Con(ZFC) is).

Because the model believes Con(ZFC) and also believes that no model of ZFC can believe Con(ZFC), the only models of ZFC it can know will be ones that it considers to have non-standard integers. Non-standard models of arithmetic are never well-founded, so such an internal model cannot be transitive.

share|cite|improve this answer

edited Aug 19 at 2:01

answered Aug 19 at 1:46

Henning Makholm

229k16295526

Any model of ZFC+Con(ZFC)+$neg$Con(ZFC+Con(ZFC)) should do. (This theory is, by the second incompleteness theorem, consistent if ZFC+Con(ZFC) is).

Because the model believes Con(ZFC) and also believes that no model of ZFC can believe Con(ZFC), the only models of ZFC it can know will be ones that it considers to have non-standard integers. Non-standard models of arithmetic are never well-founded, so such an internal model cannot be transitive.

share|cite|improve this answer

edited Aug 19 at 2:01

answered Aug 19 at 1:46

Henning Makholm

229k16295526

share|cite|improve this answer

share|cite|improve this answer

edited Aug 19 at 2:01

edited Aug 19 at 2:01

edited Aug 19 at 2:01

answered Aug 19 at 1:46

Henning Makholm

229k16295526

answered Aug 19 at 1:46

Henning Makholm

229k16295526

answered Aug 19 at 1:46

Henning Makholm

229k16295526

229k16295526

add a comment | 

add a comment | 

 

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