Is the standard model for the language of number theory elementarily equivalent to one with a nonstandard element?
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On page 89 in A Friendly Introduction to Mathematical Logic, the author writes that the standard model $mathfrakN$ for $mathcalL_NT$ is elementarily equivalent to a model $mathfrakA$ that has an element of the universe $c$ that is larger than all other numbers.
I'm new to mathematical logic, but I understand that elementarily equivalent means the two structures have the same set of true sentences. However, it seems to me that the following sentence is true in $mathfrakA$ but not in $mathfrakN$. What am I missing?
$exists x forall y (x=y vee y<x)$
logic model-theory nonstandard-models
asked Sep 27 at 13:10
Katie Johnson
434
add a comment |Â
up vote
8
down vote
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On page 89 in A Friendly Introduction to Mathematical Logic, the author writes that the standard model $mathfrakN$ for $mathcalL_NT$ is elementarily equivalent to a model $mathfrakA$ that has an element of the universe $c$ that is larger than all other numbers.
I'm new to mathematical logic, but I understand that elementarily equivalent means the two structures have the same set of true sentences. However, it seems to me that the following sentence is true in $mathfrakA$ but not in $mathfrakN$. What am I missing?
$exists x forall y (x=y vee y<x)$
logic model-theory nonstandard-models
asked Sep 27 at 13:10
Katie Johnson
434
1
The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $mathfrakA$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $mathfrakNrightarrowmathfrakA$), but this does not mean that $c$ is bigger than all elements of $mathfrakA$.
– Noah Schweber
Sep 27 at 15:41
Must the universe of $mathfrakA$ contain other nonstandard numbers?
– Katie Johnson
Sep 27 at 19:26
3
Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $mathfrakN$ exists in $mathfrakA$ as well, even for the nonstandard elements, since $mathfrakNequivmathfrakA$.
– Noah Schweber
Sep 27 at 19:28
Yes, of course. Thank you!! :)
– Katie Johnson
Sep 27 at 19:38
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
On page 89 in A Friendly Introduction to Mathematical Logic, the author writes that the standard model $mathfrakN$ for $mathcalL_NT$ is elementarily equivalent to a model $mathfrakA$ that has an element of the universe $c$ that is larger than all other numbers.
I'm new to mathematical logic, but I understand that elementarily equivalent means the two structures have the same set of true sentences. However, it seems to me that the following sentence is true in $mathfrakA$ but not in $mathfrakN$. What am I missing?
$exists x forall y (x=y vee y<x)$
logic model-theory nonstandard-models
asked Sep 27 at 13:10
Katie Johnson
434
On page 89 in A Friendly Introduction to Mathematical Logic, the author writes that the standard model $mathfrakN$ for $mathcalL_NT$ is elementarily equivalent to a model $mathfrakA$ that has an element of the universe $c$ that is larger than all other numbers.
I'm new to mathematical logic, but I understand that elementarily equivalent means the two structures have the same set of true sentences. However, it seems to me that the following sentence is true in $mathfrakA$ but not in $mathfrakN$. What am I missing?
$exists x forall y (x=y vee y<x)$
logic model-theory nonstandard-models
logic model-theory nonstandard-models
asked Sep 27 at 13:10
Katie Johnson
434
asked Sep 27 at 13:10
Katie Johnson
434
asked Sep 27 at 13:10
Katie Johnson
434
asked Sep 27 at 13:10
Katie Johnson
434
asked Sep 27 at 13:10
Katie Johnson
434
434
1
The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $mathfrakA$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $mathfrakNrightarrowmathfrakA$), but this does not mean that $c$ is bigger than all elements of $mathfrakA$.
– Noah Schweber
Sep 27 at 15:41
Must the universe of $mathfrakA$ contain other nonstandard numbers?
– Katie Johnson
Sep 27 at 19:26
3
Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $mathfrakN$ exists in $mathfrakA$ as well, even for the nonstandard elements, since $mathfrakNequivmathfrakA$.
– Noah Schweber
Sep 27 at 19:28
Yes, of course. Thank you!! :)
– Katie Johnson
Sep 27 at 19:38
add a comment |Â
1
The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $mathfrakA$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $mathfrakNrightarrowmathfrakA$), but this does not mean that $c$ is bigger than all elements of $mathfrakA$.
– Noah Schweber
Sep 27 at 15:41
Must the universe of $mathfrakA$ contain other nonstandard numbers?
– Katie Johnson
Sep 27 at 19:26
3
Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $mathfrakN$ exists in $mathfrakA$ as well, even for the nonstandard elements, since $mathfrakNequivmathfrakA$.
– Noah Schweber
Sep 27 at 19:28
Yes, of course. Thank you!! :)
– Katie Johnson
Sep 27 at 19:38
1
1
The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $mathfrakA$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $mathfrakNrightarrowmathfrakA$), but this does not mean that $c$ is bigger than all elements of $mathfrakA$.
– Noah Schweber
Sep 27 at 15:41
The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $mathfrakA$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $mathfrakNrightarrowmathfrakA$), but this does not mean that $c$ is bigger than all elements of $mathfrakA$.
– Noah Schweber
Sep 27 at 15:41
Must the universe of $mathfrakA$ contain other nonstandard numbers?
– Katie Johnson
Sep 27 at 19:26
Must the universe of $mathfrakA$ contain other nonstandard numbers?
– Katie Johnson
Sep 27 at 19:26
3
3
Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $mathfrakN$ exists in $mathfrakA$ as well, even for the nonstandard elements, since $mathfrakNequivmathfrakA$.
– Noah Schweber
Sep 27 at 19:28
Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $mathfrakN$ exists in $mathfrakA$ as well, even for the nonstandard elements, since $mathfrakNequivmathfrakA$.
– Noah Schweber
Sep 27 at 19:28
Yes, of course. Thank you!! :)
– Katie Johnson
Sep 27 at 19:38
Yes, of course. Thank you!! :)
– Katie Johnson
Sep 27 at 19:38
add a comment |Â
1 Answer
1
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up vote
7
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accepted
In the notes, I don't see the claim that $c$ is larger than all other numbers of $mathfrakA$. The number $c$ in $mathfrakA$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $mathfrakN$. But there will be other elements of $mathfrakA$ that are larger than $c$. Not every element of $mathfrakA$ is of the form $S^n(0)$ for some $n in mathfrakN$.
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
1
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
1
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
In the notes, I don't see the claim that $c$ is larger than all other numbers of $mathfrakA$. The number $c$ in $mathfrakA$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $mathfrakN$. But there will be other elements of $mathfrakA$ that are larger than $c$. Not every element of $mathfrakA$ is of the form $S^n(0)$ for some $n in mathfrakN$.
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
1
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
1
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
add a comment |Â
up vote
7
down vote
accepted
In the notes, I don't see the claim that $c$ is larger than all other numbers of $mathfrakA$. The number $c$ in $mathfrakA$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $mathfrakN$. But there will be other elements of $mathfrakA$ that are larger than $c$. Not every element of $mathfrakA$ is of the form $S^n(0)$ for some $n in mathfrakN$.
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
1
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
1
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
In the notes, I don't see the claim that $c$ is larger than all other numbers of $mathfrakA$. The number $c$ in $mathfrakA$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $mathfrakN$. But there will be other elements of $mathfrakA$ that are larger than $c$. Not every element of $mathfrakA$ is of the form $S^n(0)$ for some $n in mathfrakN$.
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
In the notes, I don't see the claim that $c$ is larger than all other numbers of $mathfrakA$. The number $c$ in $mathfrakA$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $mathfrakN$. But there will be other elements of $mathfrakA$ that are larger than $c$. Not every element of $mathfrakA$ is of the form $S^n(0)$ for some $n in mathfrakN$.
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
answered Sep 27 at 14:05
Carl Mummert
64.8k7128241
64.8k7128241
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
1
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
1
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
add a comment |Â
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
1
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
1
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
Thank you! I'm starting to understand. But can't we define $mathfrakA$ so that the universe is exactly $mathbbN cup c$? I was imagining $mathfrakA$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case.
– Katie Johnson
Sep 27 at 19:24
1
1
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
@KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $mathfrakN$. This is a very strong condition, and we can't just handwave it away.
– Noah Schweber
Sep 27 at 19:29
1
1
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
@Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc.
– Carl Mummert
Sep 27 at 20:34
add a comment |Â
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1
The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $mathfrakA$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $mathfrakNrightarrowmathfrakA$), but this does not mean that $c$ is bigger than all elements of $mathfrakA$.
– Noah Schweber
Sep 27 at 15:41
Must the universe of $mathfrakA$ contain other nonstandard numbers?
– Katie Johnson
Sep 27 at 19:26
3
Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $mathfrakN$ exists in $mathfrakA$ as well, even for the nonstandard elements, since $mathfrakNequivmathfrakA$.
– Noah Schweber
Sep 27 at 19:28
Yes, of course. Thank you!! :)
– Katie Johnson
Sep 27 at 19:38