Union of a strictly increasing sequence of closed theories in a finite language has an infinite model

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The union of a strictly increasing sequence of closed theories in a finite language has an infinite model.

This is an exercise in the book by Chang-Keisler. I don't have any idea to prove it. I just know that the union of a strictly increasing sequence of closed theories is a consistent closed theory which is not finitely axiomatizable, and also that the model of the union of consistent increasing theories is also the model of each theory.
If the union was complete the problem would be solved.

logic first-order-logic model-theory

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edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

asked Jul 15 at 15:20

user470412

44917

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: use ultraproduct.
  – Berci
  Jul 15 at 15:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ultra product? How?
  – user470412
  Jul 15 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is a "closed" theory? Does that just mean deductively closed?
  – Eric Wofsey
  Jul 15 at 17:02
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

2

The union of a strictly increasing sequence of closed theories in a finite language has an infinite model.

This is an exercise in the book by Chang-Keisler. I don't have any idea to prove it. I just know that the union of a strictly increasing sequence of closed theories is a consistent closed theory which is not finitely axiomatizable, and also that the model of the union of consistent increasing theories is also the model of each theory.
If the union was complete the problem would be solved.

logic first-order-logic model-theory

share|cite|improve this question

edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

asked Jul 15 at 15:20

user470412

44917

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: use ultraproduct.
  – Berci
  Jul 15 at 15:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ultra product? How?
  – user470412
  Jul 15 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is a "closed" theory? Does that just mean deductively closed?
  – Eric Wofsey
  Jul 15 at 17:02
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

2

up vote
4
down vote

favorite

2

2

The union of a strictly increasing sequence of closed theories in a finite language has an infinite model.

This is an exercise in the book by Chang-Keisler. I don't have any idea to prove it. I just know that the union of a strictly increasing sequence of closed theories is a consistent closed theory which is not finitely axiomatizable, and also that the model of the union of consistent increasing theories is also the model of each theory.
If the union was complete the problem would be solved.

logic first-order-logic model-theory

share|cite|improve this question

edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

asked Jul 15 at 15:20

user470412

44917

The union of a strictly increasing sequence of closed theories in a finite language has an infinite model.

This is an exercise in the book by Chang-Keisler. I don't have any idea to prove it. I just know that the union of a strictly increasing sequence of closed theories is a consistent closed theory which is not finitely axiomatizable, and also that the model of the union of consistent increasing theories is also the model of each theory.
If the union was complete the problem would be solved.

logic first-order-logic model-theory

share|cite|improve this question

edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

asked Jul 15 at 15:20

user470412

44917

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

edited Jul 15 at 16:21

Jendrik Stelzner

7,55211037

7,55211037

asked Jul 15 at 15:20

user470412

44917

asked Jul 15 at 15:20

user470412

44917

asked Jul 15 at 15:20

user470412

44917

44917

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: use ultraproduct.
  – Berci
  Jul 15 at 15:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ultra product? How?
  – user470412
  Jul 15 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is a "closed" theory? Does that just mean deductively closed?
  – Eric Wofsey
  Jul 15 at 17:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: use ultraproduct.
  – Berci
  Jul 15 at 15:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ultra product? How?
  – user470412
  Jul 15 at 15:27
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  What is a "closed" theory? Does that just mean deductively closed?
  – Eric Wofsey
  Jul 15 at 17:02
  

  
  
  
  

2

2

Hint: use ultraproduct.
– Berci
Jul 15 at 15:22

Hint: use ultraproduct.
– Berci
Jul 15 at 15:22

Ultra product? How?
– user470412
Jul 15 at 15:27

Ultra product? How?
– user470412
Jul 15 at 15:27

2

2

What is a "closed" theory? Does that just mean deductively closed?
– Eric Wofsey
Jul 15 at 17:02

What is a "closed" theory? Does that just mean deductively closed?
– Eric Wofsey
Jul 15 at 17:02

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

I don't want to give away the whole problem, but here are a couple hints to get started:

• First, show that if $Sigma$ is a finite language then for each $n$ there are only finitely many $Sigma$-structures of cardinality $le n$ (up to isomorphism). In fact, more is true: every finite $Sigma$-structure is characterized up to isomorphism by a single $Sigma$-sentence. This isn't needed here, though.




• Next, remember that by the compactness theorem, if a theory has arbitrarily large finite models then it has infinite models.




• So suppose $(T_i)_iinmathbbN$ is an increasing sequence of closed theories such that $T:=bigcup T_i$ has no infinite model. Then some $T_i$ satisfies "there are at most $n$ elements in the domain" for some $n$; in particular, by the first bulletpoint there is some finite set of structures $mathcalM_1,...,mathcalM_k$ such that every model of $T_i$ is isomorphic to one of the $mathcalM_j$s.




• Now, suppose $S_2$ is a closed theory properly containing $S_1$. What can you say about the set of models of $S_2$ versus the set of models of $S_1$? Looking back to the previous bulletpoint, what does this say about any increasing sequence $(T_i)_iinmathbbN$ of closed theories whose union has no infinite models?

share|cite|improve this answer

answered Jul 15 at 17:18

Noah Schweber

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't understand in step three why we have such a finite set ?
  – user470412
  Jul 15 at 17:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why some $T_i$ satisfies that sentence? Is the union complete?
  – user470412
  Jul 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
  – Noah Schweber
  Jul 15 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
  – Noah Schweber
  Jul 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
  – user470412
  Jul 15 at 18:00
  

  
  
  
  

 | 
show 6 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

I don't want to give away the whole problem, but here are a couple hints to get started:

• First, show that if $Sigma$ is a finite language then for each $n$ there are only finitely many $Sigma$-structures of cardinality $le n$ (up to isomorphism). In fact, more is true: every finite $Sigma$-structure is characterized up to isomorphism by a single $Sigma$-sentence. This isn't needed here, though.




• Next, remember that by the compactness theorem, if a theory has arbitrarily large finite models then it has infinite models.




• So suppose $(T_i)_iinmathbbN$ is an increasing sequence of closed theories such that $T:=bigcup T_i$ has no infinite model. Then some $T_i$ satisfies "there are at most $n$ elements in the domain" for some $n$; in particular, by the first bulletpoint there is some finite set of structures $mathcalM_1,...,mathcalM_k$ such that every model of $T_i$ is isomorphic to one of the $mathcalM_j$s.




• Now, suppose $S_2$ is a closed theory properly containing $S_1$. What can you say about the set of models of $S_2$ versus the set of models of $S_1$? Looking back to the previous bulletpoint, what does this say about any increasing sequence $(T_i)_iinmathbbN$ of closed theories whose union has no infinite models?

share|cite|improve this answer

answered Jul 15 at 17:18

Noah Schweber

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't understand in step three why we have such a finite set ?
  – user470412
  Jul 15 at 17:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why some $T_i$ satisfies that sentence? Is the union complete?
  – user470412
  Jul 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
  – Noah Schweber
  Jul 15 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
  – Noah Schweber
  Jul 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
  – user470412
  Jul 15 at 18:00
  

  
  
  
  

 | 
show 6 more comments

up vote
3
down vote



accepted

I don't want to give away the whole problem, but here are a couple hints to get started:

• First, show that if $Sigma$ is a finite language then for each $n$ there are only finitely many $Sigma$-structures of cardinality $le n$ (up to isomorphism). In fact, more is true: every finite $Sigma$-structure is characterized up to isomorphism by a single $Sigma$-sentence. This isn't needed here, though.




• Next, remember that by the compactness theorem, if a theory has arbitrarily large finite models then it has infinite models.




• So suppose $(T_i)_iinmathbbN$ is an increasing sequence of closed theories such that $T:=bigcup T_i$ has no infinite model. Then some $T_i$ satisfies "there are at most $n$ elements in the domain" for some $n$; in particular, by the first bulletpoint there is some finite set of structures $mathcalM_1,...,mathcalM_k$ such that every model of $T_i$ is isomorphic to one of the $mathcalM_j$s.




• Now, suppose $S_2$ is a closed theory properly containing $S_1$. What can you say about the set of models of $S_2$ versus the set of models of $S_1$? Looking back to the previous bulletpoint, what does this say about any increasing sequence $(T_i)_iinmathbbN$ of closed theories whose union has no infinite models?

share|cite|improve this answer

answered Jul 15 at 17:18

Noah Schweber

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't understand in step three why we have such a finite set ?
  – user470412
  Jul 15 at 17:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why some $T_i$ satisfies that sentence? Is the union complete?
  – user470412
  Jul 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
  – Noah Schweber
  Jul 15 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
  – Noah Schweber
  Jul 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
  – user470412
  Jul 15 at 18:00
  

  
  
  
  

 | 
show 6 more comments

up vote
3
down vote



accepted

up vote
3
down vote



accepted

I don't want to give away the whole problem, but here are a couple hints to get started:

• First, show that if $Sigma$ is a finite language then for each $n$ there are only finitely many $Sigma$-structures of cardinality $le n$ (up to isomorphism). In fact, more is true: every finite $Sigma$-structure is characterized up to isomorphism by a single $Sigma$-sentence. This isn't needed here, though.




• Next, remember that by the compactness theorem, if a theory has arbitrarily large finite models then it has infinite models.




• So suppose $(T_i)_iinmathbbN$ is an increasing sequence of closed theories such that $T:=bigcup T_i$ has no infinite model. Then some $T_i$ satisfies "there are at most $n$ elements in the domain" for some $n$; in particular, by the first bulletpoint there is some finite set of structures $mathcalM_1,...,mathcalM_k$ such that every model of $T_i$ is isomorphic to one of the $mathcalM_j$s.




• Now, suppose $S_2$ is a closed theory properly containing $S_1$. What can you say about the set of models of $S_2$ versus the set of models of $S_1$? Looking back to the previous bulletpoint, what does this say about any increasing sequence $(T_i)_iinmathbbN$ of closed theories whose union has no infinite models?

share|cite|improve this answer

answered Jul 15 at 17:18

Noah Schweber

111k9140264

I don't want to give away the whole problem, but here are a couple hints to get started:

• First, show that if $Sigma$ is a finite language then for each $n$ there are only finitely many $Sigma$-structures of cardinality $le n$ (up to isomorphism). In fact, more is true: every finite $Sigma$-structure is characterized up to isomorphism by a single $Sigma$-sentence. This isn't needed here, though.




• Next, remember that by the compactness theorem, if a theory has arbitrarily large finite models then it has infinite models.




• So suppose $(T_i)_iinmathbbN$ is an increasing sequence of closed theories such that $T:=bigcup T_i$ has no infinite model. Then some $T_i$ satisfies "there are at most $n$ elements in the domain" for some $n$; in particular, by the first bulletpoint there is some finite set of structures $mathcalM_1,...,mathcalM_k$ such that every model of $T_i$ is isomorphic to one of the $mathcalM_j$s.




• Now, suppose $S_2$ is a closed theory properly containing $S_1$. What can you say about the set of models of $S_2$ versus the set of models of $S_1$? Looking back to the previous bulletpoint, what does this say about any increasing sequence $(T_i)_iinmathbbN$ of closed theories whose union has no infinite models?

share|cite|improve this answer

answered Jul 15 at 17:18

Noah Schweber

111k9140264

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 17:18

Noah Schweber

111k9140264

answered Jul 15 at 17:18

Noah Schweber

111k9140264

answered Jul 15 at 17:18

Noah Schweber

111k9140264

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't understand in step three why we have such a finite set ?
  – user470412
  Jul 15 at 17:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why some $T_i$ satisfies that sentence? Is the union complete?
  – user470412
  Jul 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
  – Noah Schweber
  Jul 15 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
  – Noah Schweber
  Jul 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
  – user470412
  Jul 15 at 18:00
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't understand in step three why we have such a finite set ?
  – user470412
  Jul 15 at 17:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why some $T_i$ satisfies that sentence? Is the union complete?
  – user470412
  Jul 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
  – Noah Schweber
  Jul 15 at 17:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
  – Noah Schweber
  Jul 15 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
  – user470412
  Jul 15 at 18:00
  

  
  
  
  

I can't understand in step three why we have such a finite set ?
– user470412
Jul 15 at 17:48

I can't understand in step three why we have such a finite set ?
– user470412
Jul 15 at 17:48

And why some $T_i$ satisfies that sentence? Is the union complete?
– user470412
Jul 15 at 17:51

And why some $T_i$ satisfies that sentence? Is the union complete?
– user470412
Jul 15 at 17:51

@user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
– Noah Schweber
Jul 15 at 17:55

@user470412 Re: your first comment, that's exactly what the first bulletpoint says: if $T_i$ satisfies "there are at most $n$ elements in the domain," then every model of $T_i$ has cardinality $le n$, and by the first bulletpoint there are only finitely many such structures (up to isomorphism).
– Noah Schweber
Jul 15 at 17:55

Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
– Noah Schweber
Jul 15 at 17:56

Re: your second comment, no, the theory is not assumed to be complete. However, by the second bulletpoint we know that if the union has arbitrarily large models it has infinite models. Since we're assuming the union has no infinite models, this means there is a bound on the size of the finite models - that is (since $T$, being a union of a chain of closed theories, is closed), for some $n$ the sentence "there are at most $n$ elements in the domain" is in $T$. Now do you see why anything in $T$ is also in one of the $T_i$s?
– Noah Schweber
Jul 15 at 17:56

In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
– user470412
Jul 15 at 18:00

In step2 you mean every model of $S_2$ is a model of $S_1$ ? And every model of the union is a model of each $T_i$, so each $T_i$ has at least one finite model. I don't know how to use this....
– user470412
Jul 15 at 18:00

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