Elementary equivalence of standard and non-standard model of arithmetic
$begingroup$
There is the common construction of a non-standard model of arithmetic by adding a constant symbol c to the signature and adding ${n<c:nin mathbb{N}}$ to the theory PA. Now by adding all sentences $sigma$ st. $mathbb{N}models sigma$ we get a theory $T^*$, which is consistent by compactness theorem. My textbook says now that any model of $T^*$ is a model of PA which is elementary equivalent to $mathbb{N}$.
However, I found definitions of elementary equivalence only for models of the same language. The language of the model constructed above differs from the standard language $mathcal{L}_{PA}$ by c.
Where am I going wrong?
model-theory nonstandard-models
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
$endgroup$
add a comment |
$begingroup$
There is the common construction of a non-standard model of arithmetic by adding a constant symbol c to the signature and adding ${n<c:nin mathbb{N}}$ to the theory PA. Now by adding all sentences $sigma$ st. $mathbb{N}models sigma$ we get a theory $T^*$, which is consistent by compactness theorem. My textbook says now that any model of $T^*$ is a model of PA which is elementary equivalent to $mathbb{N}$.
However, I found definitions of elementary equivalence only for models of the same language. The language of the model constructed above differs from the standard language $mathcal{L}_{PA}$ by c.
Where am I going wrong?
model-theory nonstandard-models
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
$endgroup$
add a comment |
$begingroup$
There is the common construction of a non-standard model of arithmetic by adding a constant symbol c to the signature and adding ${n<c:nin mathbb{N}}$ to the theory PA. Now by adding all sentences $sigma$ st. $mathbb{N}models sigma$ we get a theory $T^*$, which is consistent by compactness theorem. My textbook says now that any model of $T^*$ is a model of PA which is elementary equivalent to $mathbb{N}$.
However, I found definitions of elementary equivalence only for models of the same language. The language of the model constructed above differs from the standard language $mathcal{L}_{PA}$ by c.
Where am I going wrong?
model-theory nonstandard-models
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
$endgroup$
There is the common construction of a non-standard model of arithmetic by adding a constant symbol c to the signature and adding ${n<c:nin mathbb{N}}$ to the theory PA. Now by adding all sentences $sigma$ st. $mathbb{N}models sigma$ we get a theory $T^*$, which is consistent by compactness theorem. My textbook says now that any model of $T^*$ is a model of PA which is elementary equivalent to $mathbb{N}$.
However, I found definitions of elementary equivalence only for models of the same language. The language of the model constructed above differs from the standard language $mathcal{L}_{PA}$ by c.
Where am I going wrong?
model-theory nonstandard-models
model-theory nonstandard-models
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
asked Jan 10 at 23:56
GottlobtFregeGottlobtFrege
778
778
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any model of $T^*$ can also be considered as a model of PA by just forgetting about $c$. In other words, you consider it as a structure with only the arithmetic operations and not the constant $c$.
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
2
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
2
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
2
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
1
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
|
show 2 more comments
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Any model of $T^*$ can also be considered as a model of PA by just forgetting about $c$. In other words, you consider it as a structure with only the arithmetic operations and not the constant $c$.
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
2
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
2
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
2
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
1
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
|
show 2 more comments
$begingroup$
Any model of $T^*$ can also be considered as a model of PA by just forgetting about $c$. In other words, you consider it as a structure with only the arithmetic operations and not the constant $c$.
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
2
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
2
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
2
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
1
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
|
show 2 more comments
$begingroup$
Any model of $T^*$ can also be considered as a model of PA by just forgetting about $c$. In other words, you consider it as a structure with only the arithmetic operations and not the constant $c$.
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
$endgroup$
Any model of $T^*$ can also be considered as a model of PA by just forgetting about $c$. In other words, you consider it as a structure with only the arithmetic operations and not the constant $c$.
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
answered Jan 11 at 0:02
Eric WofseyEric Wofsey
188k14216346
188k14216346
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
2
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
2
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
2
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
1
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
|
show 2 more comments
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
2
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
2
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
2
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
1
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
$begingroup$
So elementary equivalence (and isomorphy) of models are always relative to a signature?
$endgroup$
– GottlobtFrege
Jan 11 at 0:03
2
2
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
$begingroup$
Yes, though the signature usually is not stated explicitly.
$endgroup$
– Eric Wofsey
Jan 11 at 0:04
2
2
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
$begingroup$
Of course we can still talk about the order-type, since $<$ is still part of the signature.
$endgroup$
– Eric Wofsey
Jan 11 at 0:09
2
2
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
$begingroup$
@GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $mathbb{N}$) is still there.
$endgroup$
– Noah Schweber
Jan 11 at 0:14
1
1
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
$begingroup$
@GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(mathbb N,+,cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension).
$endgroup$
– spaceisdarkgreen
Jan 11 at 2:03
|
show 2 more comments
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