How to show a class of structures is not axiomatizable?
For example, let $F$ be a field and $L$ be the language of $F$-vector space.
(1) Prove that the class of finite dimensional $F$-vector space is not axiomatizable.
(2) Prove that if $F$ is infinite then the class of infinite dimensional $F$-vector space is not axiomatizable.
Or let $L$ be the language of rings.
(3) Prove that the class of algebraic extensions of $mathbb{Q}$ is not axiomatizable.
I think the common way to prove this type of statement would be: first suppose the class is axiomatizable. Then there is some $L$-theory axiomatizing the class. We make a new language $L'$ by adding new symbols to $L$ then construct a $L'$-theory $T'$ and show $T'$ is consistent by compactness then observe a contradiction.
However, this method requires a bit of algebra knowledge. Can anyone give some hints for the above problems I have listed?
linear-algebra abstract-algebra logic model-theory
asked Nov 28 at 7:32
bbw
47038
add a comment |
For example, let $F$ be a field and $L$ be the language of $F$-vector space.
(1) Prove that the class of finite dimensional $F$-vector space is not axiomatizable.
(2) Prove that if $F$ is infinite then the class of infinite dimensional $F$-vector space is not axiomatizable.
Or let $L$ be the language of rings.
(3) Prove that the class of algebraic extensions of $mathbb{Q}$ is not axiomatizable.
I think the common way to prove this type of statement would be: first suppose the class is axiomatizable. Then there is some $L$-theory axiomatizing the class. We make a new language $L'$ by adding new symbols to $L$ then construct a $L'$-theory $T'$ and show $T'$ is consistent by compactness then observe a contradiction.
However, this method requires a bit of algebra knowledge. Can anyone give some hints for the above problems I have listed?
linear-algebra abstract-algebra logic model-theory
asked Nov 28 at 7:32
bbw
47038
add a comment |
For example, let $F$ be a field and $L$ be the language of $F$-vector space.
(1) Prove that the class of finite dimensional $F$-vector space is not axiomatizable.
(2) Prove that if $F$ is infinite then the class of infinite dimensional $F$-vector space is not axiomatizable.
Or let $L$ be the language of rings.
(3) Prove that the class of algebraic extensions of $mathbb{Q}$ is not axiomatizable.
I think the common way to prove this type of statement would be: first suppose the class is axiomatizable. Then there is some $L$-theory axiomatizing the class. We make a new language $L'$ by adding new symbols to $L$ then construct a $L'$-theory $T'$ and show $T'$ is consistent by compactness then observe a contradiction.
However, this method requires a bit of algebra knowledge. Can anyone give some hints for the above problems I have listed?
linear-algebra abstract-algebra logic model-theory
asked Nov 28 at 7:32
bbw
47038
For example, let $F$ be a field and $L$ be the language of $F$-vector space.
(1) Prove that the class of finite dimensional $F$-vector space is not axiomatizable.
(2) Prove that if $F$ is infinite then the class of infinite dimensional $F$-vector space is not axiomatizable.
Or let $L$ be the language of rings.
(3) Prove that the class of algebraic extensions of $mathbb{Q}$ is not axiomatizable.
I think the common way to prove this type of statement would be: first suppose the class is axiomatizable. Then there is some $L$-theory axiomatizing the class. We make a new language $L'$ by adding new symbols to $L$ then construct a $L'$-theory $T'$ and show $T'$ is consistent by compactness then observe a contradiction.
However, this method requires a bit of algebra knowledge. Can anyone give some hints for the above problems I have listed?
linear-algebra abstract-algebra logic model-theory
linear-algebra abstract-algebra logic model-theory
asked Nov 28 at 7:32
bbw
47038
asked Nov 28 at 7:32
bbw
47038
asked Nov 28 at 7:32
bbw
47038
asked Nov 28 at 7:32
bbw
47038
asked Nov 28 at 7:32
bbw
47038
47038
add a comment |
add a comment |
1 Answer
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Hint : for all these questions, you can use the ascending Löwenheim-Skolem theorem.
answered Nov 28 at 14:08
Max
12.6k11040
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
|
show 9 more comments
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Hint : for all these questions, you can use the ascending Löwenheim-Skolem theorem.
answered Nov 28 at 14:08
Max
12.6k11040
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
|
show 9 more comments
Hint : for all these questions, you can use the ascending Löwenheim-Skolem theorem.
answered Nov 28 at 14:08
Max
12.6k11040
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
|
show 9 more comments
Hint : for all these questions, you can use the ascending Löwenheim-Skolem theorem.
answered Nov 28 at 14:08
Max
12.6k11040
Hint : for all these questions, you can use the ascending Löwenheim-Skolem theorem.
answered Nov 28 at 14:08
Max
12.6k11040
answered Nov 28 at 14:08
Max
12.6k11040
answered Nov 28 at 14:08
Max
12.6k11040
answered Nov 28 at 14:08
Max
12.6k11040
12.6k11040
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
|
show 9 more comments
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Yes, I solved (1) by showing that if it is axiomatizable then it would contradict the LS theorem. However, for (2), I tried to play the same trick but I found it is totally consistent with LS theorem. For (3), no idea what to do.
– bbw
Nov 28 at 20:58
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
Actually you can use LS for (2) as well by using this : prove that there is a infinite dimensional vector space that is elementarily equivalent to $F$. For (3), notice that an algebraic extension of $mathbb{Q}$ must be countable
– Max
Nov 28 at 21:33
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
For (2), since $F$ is infinite, then any vector spaces with the same cardinality bigger than $F$ are isomorphic. So if it is axiomatizable then the theory is complete. So any two models are elementarily equivalent. But F itself is not a model since it is one dimensional over itself. How I can show $F$ is equivalent to an infinite dimensional vector space?
– bbw
Nov 28 at 22:19
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
By using Löwenheim-Skolem !
– Max
Nov 28 at 22:20
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
I still don't get it... the LS theorem requires an infinite $L$-structure in the hypothesis. If $F$ an infinite dimensional $F$-vector space then there will be a elementary extension of $F$ of cardinality at least $|F|$ hence they are elementary equivalent. However, the thing is, $F$ is not infinite dimensional.
– bbw
Nov 28 at 22:34
|
show 9 more comments
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