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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 3 hours ago
Katie Johnson
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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 3 hours ago
Katie Johnson
262
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
43 mins ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
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 3 hours ago
Katie Johnson
262
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 3 hours ago
Katie Johnson
262
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Katie Johnson
262
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Katie Johnson
262
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Katie Johnson
262
asked 3 hours ago
Katie Johnson
262
262
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Katie Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
43 mins ago
add a comment |Â
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
43 mins ago
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
43 mins ago
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
43 mins ago
add a comment |Â
1 Answer
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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 2 hours ago
Carl Mummert
64.6k7128240
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
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 2 hours ago
Carl Mummert
64.6k7128240
add a comment |Â
up vote
6
down vote
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 2 hours ago
Carl Mummert
64.6k7128240
add a comment |Â
up vote
6
down vote
up vote
6
down vote
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 2 hours ago
Carl Mummert
64.6k7128240
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 2 hours ago
Carl Mummert
64.6k7128240
answered 2 hours ago
Carl Mummert
64.6k7128240
answered 2 hours ago
Carl Mummert
64.6k7128240
answered 2 hours ago
Carl Mummert
64.6k7128240
64.6k7128240
add a comment |Â
add a comment |Â
Katie Johnson is a new contributor. Be nice, and check out our Code of Conduct.
Katie Johnson is a new contributor. Be nice, and check out our Code of Conduct.
Katie Johnson is a new contributor. Be nice, and check out our Code of Conduct.
Katie Johnson is a new contributor. Be nice, and check out our Code of Conduct.
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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
43 mins ago