Intuitive understanding of definitional expansion in model theory
$begingroup$
I was trying to understand what definitional expansion mean. Consider having two L-structures $mathcal A$ and $mathcal A'$ where $A'$ is an L'-expansion of $mathcal A$. We way that $mathcal A'$ is a definitional expansion of $mathcal A$ if for each symbol $s in L' setminus L$ the interpretation of $s^{mathcal A'}$ of $s$ in $mathcal A'$ is 0-definable in $mathcal A$.
I understand what zero definable for a set is, which means:
$$ varphi^{mathcal A} = { (a_1,...,a_m) in A^m mid mathcal A models varphi(a_1,...,a_m)}$$
when its zero definable the L-formula $varphi$ is only in terms of $L$ (and not the names of the elements of the underlying set $a in A$).
However in this context we have a single element. We say its zero definable ok perhaps that means the set is of size 1. Since $mathcal A'$ is an L'-extension it retains the old interpretations of symbols of $L$, fine but then it interprets new symbols to elements of the set $A$ (since both L-structures only differ in the language L vs L'). So is this just saying that the new symbols we are adding are just expressable in the original language L? Essentially they are the special singleton elements depending only in the language L?
Is that all? I think perhaps the 0-definable thing confused me and it was unclear if I really got it.
logic first-order-logic model-theory
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
$endgroup$
add a comment |
$begingroup$
I was trying to understand what definitional expansion mean. Consider having two L-structures $mathcal A$ and $mathcal A'$ where $A'$ is an L'-expansion of $mathcal A$. We way that $mathcal A'$ is a definitional expansion of $mathcal A$ if for each symbol $s in L' setminus L$ the interpretation of $s^{mathcal A'}$ of $s$ in $mathcal A'$ is 0-definable in $mathcal A$.
I understand what zero definable for a set is, which means:
$$ varphi^{mathcal A} = { (a_1,...,a_m) in A^m mid mathcal A models varphi(a_1,...,a_m)}$$
when its zero definable the L-formula $varphi$ is only in terms of $L$ (and not the names of the elements of the underlying set $a in A$).
However in this context we have a single element. We say its zero definable ok perhaps that means the set is of size 1. Since $mathcal A'$ is an L'-extension it retains the old interpretations of symbols of $L$, fine but then it interprets new symbols to elements of the set $A$ (since both L-structures only differ in the language L vs L'). So is this just saying that the new symbols we are adding are just expressable in the original language L? Essentially they are the special singleton elements depending only in the language L?
Is that all? I think perhaps the 0-definable thing confused me and it was unclear if I really got it.
logic first-order-logic model-theory
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
$endgroup$
1
$begingroup$
I'm not understanding what the "single element" you're talking about is. Are you asking about the case where $s$ is a constant symbol (as opposed to a relation symbol)?
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:33
add a comment |
$begingroup$
I was trying to understand what definitional expansion mean. Consider having two L-structures $mathcal A$ and $mathcal A'$ where $A'$ is an L'-expansion of $mathcal A$. We way that $mathcal A'$ is a definitional expansion of $mathcal A$ if for each symbol $s in L' setminus L$ the interpretation of $s^{mathcal A'}$ of $s$ in $mathcal A'$ is 0-definable in $mathcal A$.
I understand what zero definable for a set is, which means:
$$ varphi^{mathcal A} = { (a_1,...,a_m) in A^m mid mathcal A models varphi(a_1,...,a_m)}$$
when its zero definable the L-formula $varphi$ is only in terms of $L$ (and not the names of the elements of the underlying set $a in A$).
However in this context we have a single element. We say its zero definable ok perhaps that means the set is of size 1. Since $mathcal A'$ is an L'-extension it retains the old interpretations of symbols of $L$, fine but then it interprets new symbols to elements of the set $A$ (since both L-structures only differ in the language L vs L'). So is this just saying that the new symbols we are adding are just expressable in the original language L? Essentially they are the special singleton elements depending only in the language L?
Is that all? I think perhaps the 0-definable thing confused me and it was unclear if I really got it.
logic first-order-logic model-theory
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
$endgroup$
I was trying to understand what definitional expansion mean. Consider having two L-structures $mathcal A$ and $mathcal A'$ where $A'$ is an L'-expansion of $mathcal A$. We way that $mathcal A'$ is a definitional expansion of $mathcal A$ if for each symbol $s in L' setminus L$ the interpretation of $s^{mathcal A'}$ of $s$ in $mathcal A'$ is 0-definable in $mathcal A$.
I understand what zero definable for a set is, which means:
$$ varphi^{mathcal A} = { (a_1,...,a_m) in A^m mid mathcal A models varphi(a_1,...,a_m)}$$
when its zero definable the L-formula $varphi$ is only in terms of $L$ (and not the names of the elements of the underlying set $a in A$).
However in this context we have a single element. We say its zero definable ok perhaps that means the set is of size 1. Since $mathcal A'$ is an L'-extension it retains the old interpretations of symbols of $L$, fine but then it interprets new symbols to elements of the set $A$ (since both L-structures only differ in the language L vs L'). So is this just saying that the new symbols we are adding are just expressable in the original language L? Essentially they are the special singleton elements depending only in the language L?
Is that all? I think perhaps the 0-definable thing confused me and it was unclear if I really got it.
logic first-order-logic model-theory
logic first-order-logic model-theory
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 4:12
PinocchioPinocchio
1,90521854
1,90521854
1
$begingroup$
I'm not understanding what the "single element" you're talking about is. Are you asking about the case where $s$ is a constant symbol (as opposed to a relation symbol)?
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:33
add a comment |
1
$begingroup$
I'm not understanding what the "single element" you're talking about is. Are you asking about the case where $s$ is a constant symbol (as opposed to a relation symbol)?
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:33
1
1
$begingroup$
I'm not understanding what the "single element" you're talking about is. Are you asking about the case where $s$ is a constant symbol (as opposed to a relation symbol)?
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:33
$begingroup$
I'm not understanding what the "single element" you're talking about is. Are you asking about the case where $s$ is a constant symbol (as opposed to a relation symbol)?
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A simple example should help clear this up. Let $L = { cdot }$, the language with just a single binary function, and let $(G, cdot)$ be a group. I claim that the identity element of $G$ is a 0-definable element. Let $phi(x)$ be the formula $forall y (x cdot y = y wedge y cdot x = y)$. Then clearly, if $e$ is the identity element, then $e$ is the unique element of $G$ such that $G models phi(e)$.
Hence, $(G, cdot, e)$ is a definition expansion of $(G, cdot)$.
We should note, though, that expansions by constants are not the only kinds of definitional expansions. For example, in a group, one could look at the center of the group: $Z(G) = { x in G : forall y (xy = yx) }$. This is a zero-definable subset of $G$, and so the expansion $(G, cdot, Z(G))$ is a definitional expansion of $G$ (in a language $L^prime$ which contains one extra unary relational symbol).
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
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add a comment |
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$begingroup$
A simple example should help clear this up. Let $L = { cdot }$, the language with just a single binary function, and let $(G, cdot)$ be a group. I claim that the identity element of $G$ is a 0-definable element. Let $phi(x)$ be the formula $forall y (x cdot y = y wedge y cdot x = y)$. Then clearly, if $e$ is the identity element, then $e$ is the unique element of $G$ such that $G models phi(e)$.
Hence, $(G, cdot, e)$ is a definition expansion of $(G, cdot)$.
We should note, though, that expansions by constants are not the only kinds of definitional expansions. For example, in a group, one could look at the center of the group: $Z(G) = { x in G : forall y (xy = yx) }$. This is a zero-definable subset of $G$, and so the expansion $(G, cdot, Z(G))$ is a definitional expansion of $G$ (in a language $L^prime$ which contains one extra unary relational symbol).
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
$endgroup$
add a comment |
$begingroup$
A simple example should help clear this up. Let $L = { cdot }$, the language with just a single binary function, and let $(G, cdot)$ be a group. I claim that the identity element of $G$ is a 0-definable element. Let $phi(x)$ be the formula $forall y (x cdot y = y wedge y cdot x = y)$. Then clearly, if $e$ is the identity element, then $e$ is the unique element of $G$ such that $G models phi(e)$.
Hence, $(G, cdot, e)$ is a definition expansion of $(G, cdot)$.
We should note, though, that expansions by constants are not the only kinds of definitional expansions. For example, in a group, one could look at the center of the group: $Z(G) = { x in G : forall y (xy = yx) }$. This is a zero-definable subset of $G$, and so the expansion $(G, cdot, Z(G))$ is a definitional expansion of $G$ (in a language $L^prime$ which contains one extra unary relational symbol).
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
$endgroup$
add a comment |
$begingroup$
A simple example should help clear this up. Let $L = { cdot }$, the language with just a single binary function, and let $(G, cdot)$ be a group. I claim that the identity element of $G$ is a 0-definable element. Let $phi(x)$ be the formula $forall y (x cdot y = y wedge y cdot x = y)$. Then clearly, if $e$ is the identity element, then $e$ is the unique element of $G$ such that $G models phi(e)$.
Hence, $(G, cdot, e)$ is a definition expansion of $(G, cdot)$.
We should note, though, that expansions by constants are not the only kinds of definitional expansions. For example, in a group, one could look at the center of the group: $Z(G) = { x in G : forall y (xy = yx) }$. This is a zero-definable subset of $G$, and so the expansion $(G, cdot, Z(G))$ is a definitional expansion of $G$ (in a language $L^prime$ which contains one extra unary relational symbol).
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
$endgroup$
A simple example should help clear this up. Let $L = { cdot }$, the language with just a single binary function, and let $(G, cdot)$ be a group. I claim that the identity element of $G$ is a 0-definable element. Let $phi(x)$ be the formula $forall y (x cdot y = y wedge y cdot x = y)$. Then clearly, if $e$ is the identity element, then $e$ is the unique element of $G$ such that $G models phi(e)$.
Hence, $(G, cdot, e)$ is a definition expansion of $(G, cdot)$.
We should note, though, that expansions by constants are not the only kinds of definitional expansions. For example, in a group, one could look at the center of the group: $Z(G) = { x in G : forall y (xy = yx) }$. This is a zero-definable subset of $G$, and so the expansion $(G, cdot, Z(G))$ is a definitional expansion of $G$ (in a language $L^prime$ which contains one extra unary relational symbol).
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
answered Dec 19 '18 at 13:41
Athar Abdul-QuaderAthar Abdul-Quader
81748
81748
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$begingroup$
I'm not understanding what the "single element" you're talking about is. Are you asking about the case where $s$ is a constant symbol (as opposed to a relation symbol)?
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:33