Reduct of a first-order structure
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I am trying to understand the notion of reduct in model theory. I found the following definition in Hodges's model theory book.
Definition. Let $mathcal{L}$ be a first-order language and $mathcal{L}'subseteq mathcal{L}$. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure. A structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if $mathcal{N}=mathcal{M}|_{mathcal{L}'}$.
This definition is clear for me. We just restrict $mathcal{M}$ to $mathcal{L}'$ and we get a reduct. But if you google reduct you will find several papers in which there are some difference between their definition of reduct. For example in one of them I found the following definition.
Definition. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure where $mathcal{L}$ is a relational language. An $mathcal{L}'$-structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if every relation in $mathcal{L}'$ is 0-definable in $mathcal{L}$.
Could you explain why these definitions are the same?
logic first-order-logic model-theory
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
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add a comment |
$begingroup$
I am trying to understand the notion of reduct in model theory. I found the following definition in Hodges's model theory book.
Definition. Let $mathcal{L}$ be a first-order language and $mathcal{L}'subseteq mathcal{L}$. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure. A structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if $mathcal{N}=mathcal{M}|_{mathcal{L}'}$.
This definition is clear for me. We just restrict $mathcal{M}$ to $mathcal{L}'$ and we get a reduct. But if you google reduct you will find several papers in which there are some difference between their definition of reduct. For example in one of them I found the following definition.
Definition. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure where $mathcal{L}$ is a relational language. An $mathcal{L}'$-structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if every relation in $mathcal{L}'$ is 0-definable in $mathcal{L}$.
Could you explain why these definitions are the same?
logic first-order-logic model-theory
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
$endgroup$
1
$begingroup$
Why do you believe the definition of "reduct" will be identical in similar (but very slightly different) settings?
$endgroup$
– Eric Towers
Dec 28 '18 at 2:31
add a comment |
$begingroup$
I am trying to understand the notion of reduct in model theory. I found the following definition in Hodges's model theory book.
Definition. Let $mathcal{L}$ be a first-order language and $mathcal{L}'subseteq mathcal{L}$. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure. A structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if $mathcal{N}=mathcal{M}|_{mathcal{L}'}$.
This definition is clear for me. We just restrict $mathcal{M}$ to $mathcal{L}'$ and we get a reduct. But if you google reduct you will find several papers in which there are some difference between their definition of reduct. For example in one of them I found the following definition.
Definition. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure where $mathcal{L}$ is a relational language. An $mathcal{L}'$-structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if every relation in $mathcal{L}'$ is 0-definable in $mathcal{L}$.
Could you explain why these definitions are the same?
logic first-order-logic model-theory
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
$endgroup$
I am trying to understand the notion of reduct in model theory. I found the following definition in Hodges's model theory book.
Definition. Let $mathcal{L}$ be a first-order language and $mathcal{L}'subseteq mathcal{L}$. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure. A structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if $mathcal{N}=mathcal{M}|_{mathcal{L}'}$.
This definition is clear for me. We just restrict $mathcal{M}$ to $mathcal{L}'$ and we get a reduct. But if you google reduct you will find several papers in which there are some difference between their definition of reduct. For example in one of them I found the following definition.
Definition. Let $mathcal{M}=(M,dots)$ be an $mathcal{L}$-structure where $mathcal{L}$ is a relational language. An $mathcal{L}'$-structure $mathcal{N}=(M,dots)$ is called a reduct of $mathcal{M}$ if every relation in $mathcal{L}'$ is 0-definable in $mathcal{L}$.
Could you explain why these definitions are the same?
logic first-order-logic model-theory
logic first-order-logic model-theory
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
asked Dec 28 '18 at 2:22
Alice.HAlice.H
1067
1067
1
$begingroup$
Why do you believe the definition of "reduct" will be identical in similar (but very slightly different) settings?
$endgroup$
– Eric Towers
Dec 28 '18 at 2:31
add a comment |
1
$begingroup$
Why do you believe the definition of "reduct" will be identical in similar (but very slightly different) settings?
$endgroup$
– Eric Towers
Dec 28 '18 at 2:31
1
1
$begingroup$
Why do you believe the definition of "reduct" will be identical in similar (but very slightly different) settings?
$endgroup$
– Eric Towers
Dec 28 '18 at 2:31
$begingroup$
Why do you believe the definition of "reduct" will be identical in similar (but very slightly different) settings?
$endgroup$
– Eric Towers
Dec 28 '18 at 2:31
add a comment |
1 Answer
1
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oldest
votes
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This is a bit of terminology I've always found annoying myself. The two definitions of "reduct" are not the same, although they are closely related. To distinguish them, I'll refer to the first as "reduct$_1$" and the second as "reduct$_2$."
The relation between them is this:
$mathcal{M}$ is a reduct$_2$ of $mathcal{N}$ iff $mathcal{M}$ is a reduct$_1$ of some (parameter-free) definitional expansion of $mathcal{N}$.
That is, to get a reduct$_2$, we take the starting structure, possibly add some additional relations/functions/constants, but only in a $(emptyset-)$definable way, and then remove some of the symbols. Every reduct$_1$ is a reduct$_2$, but the converse fails quite badly: e.g. $(mathbb{N}, <)$ is a reduct$_2$, but not a reduct$_1$, of $(mathbb{N},+)$.
The motivation for the terminology overload is that from a model-theoretic perspective, there is little difference between the "primitive" relations/functions/constants and the definable relations/functions/constants. For example, taking definitional expansions preserves: decidability, saturation, homogeneity, minimality and its various cousins, stability, ... So much of the time in model theory we lose nothing by conflating a given structure with its "full definitional expansion" (add all the $emptyset-$definable relations/functions/constants), and then the two notions of "reduct" essentially coincide.
But that said, it is still an overload of terminology and can definitely lead to confusion if one is not careful. That's unfortunate, but that's the situation.
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
$endgroup$
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
add a comment |
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$begingroup$
This is a bit of terminology I've always found annoying myself. The two definitions of "reduct" are not the same, although they are closely related. To distinguish them, I'll refer to the first as "reduct$_1$" and the second as "reduct$_2$."
The relation between them is this:
$mathcal{M}$ is a reduct$_2$ of $mathcal{N}$ iff $mathcal{M}$ is a reduct$_1$ of some (parameter-free) definitional expansion of $mathcal{N}$.
That is, to get a reduct$_2$, we take the starting structure, possibly add some additional relations/functions/constants, but only in a $(emptyset-)$definable way, and then remove some of the symbols. Every reduct$_1$ is a reduct$_2$, but the converse fails quite badly: e.g. $(mathbb{N}, <)$ is a reduct$_2$, but not a reduct$_1$, of $(mathbb{N},+)$.
The motivation for the terminology overload is that from a model-theoretic perspective, there is little difference between the "primitive" relations/functions/constants and the definable relations/functions/constants. For example, taking definitional expansions preserves: decidability, saturation, homogeneity, minimality and its various cousins, stability, ... So much of the time in model theory we lose nothing by conflating a given structure with its "full definitional expansion" (add all the $emptyset-$definable relations/functions/constants), and then the two notions of "reduct" essentially coincide.
But that said, it is still an overload of terminology and can definitely lead to confusion if one is not careful. That's unfortunate, but that's the situation.
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
$endgroup$
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
add a comment |
$begingroup$
This is a bit of terminology I've always found annoying myself. The two definitions of "reduct" are not the same, although they are closely related. To distinguish them, I'll refer to the first as "reduct$_1$" and the second as "reduct$_2$."
The relation between them is this:
$mathcal{M}$ is a reduct$_2$ of $mathcal{N}$ iff $mathcal{M}$ is a reduct$_1$ of some (parameter-free) definitional expansion of $mathcal{N}$.
That is, to get a reduct$_2$, we take the starting structure, possibly add some additional relations/functions/constants, but only in a $(emptyset-)$definable way, and then remove some of the symbols. Every reduct$_1$ is a reduct$_2$, but the converse fails quite badly: e.g. $(mathbb{N}, <)$ is a reduct$_2$, but not a reduct$_1$, of $(mathbb{N},+)$.
The motivation for the terminology overload is that from a model-theoretic perspective, there is little difference between the "primitive" relations/functions/constants and the definable relations/functions/constants. For example, taking definitional expansions preserves: decidability, saturation, homogeneity, minimality and its various cousins, stability, ... So much of the time in model theory we lose nothing by conflating a given structure with its "full definitional expansion" (add all the $emptyset-$definable relations/functions/constants), and then the two notions of "reduct" essentially coincide.
But that said, it is still an overload of terminology and can definitely lead to confusion if one is not careful. That's unfortunate, but that's the situation.
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
$endgroup$
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
add a comment |
$begingroup$
This is a bit of terminology I've always found annoying myself. The two definitions of "reduct" are not the same, although they are closely related. To distinguish them, I'll refer to the first as "reduct$_1$" and the second as "reduct$_2$."
The relation between them is this:
$mathcal{M}$ is a reduct$_2$ of $mathcal{N}$ iff $mathcal{M}$ is a reduct$_1$ of some (parameter-free) definitional expansion of $mathcal{N}$.
That is, to get a reduct$_2$, we take the starting structure, possibly add some additional relations/functions/constants, but only in a $(emptyset-)$definable way, and then remove some of the symbols. Every reduct$_1$ is a reduct$_2$, but the converse fails quite badly: e.g. $(mathbb{N}, <)$ is a reduct$_2$, but not a reduct$_1$, of $(mathbb{N},+)$.
The motivation for the terminology overload is that from a model-theoretic perspective, there is little difference between the "primitive" relations/functions/constants and the definable relations/functions/constants. For example, taking definitional expansions preserves: decidability, saturation, homogeneity, minimality and its various cousins, stability, ... So much of the time in model theory we lose nothing by conflating a given structure with its "full definitional expansion" (add all the $emptyset-$definable relations/functions/constants), and then the two notions of "reduct" essentially coincide.
But that said, it is still an overload of terminology and can definitely lead to confusion if one is not careful. That's unfortunate, but that's the situation.
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
$endgroup$
This is a bit of terminology I've always found annoying myself. The two definitions of "reduct" are not the same, although they are closely related. To distinguish them, I'll refer to the first as "reduct$_1$" and the second as "reduct$_2$."
The relation between them is this:
$mathcal{M}$ is a reduct$_2$ of $mathcal{N}$ iff $mathcal{M}$ is a reduct$_1$ of some (parameter-free) definitional expansion of $mathcal{N}$.
That is, to get a reduct$_2$, we take the starting structure, possibly add some additional relations/functions/constants, but only in a $(emptyset-)$definable way, and then remove some of the symbols. Every reduct$_1$ is a reduct$_2$, but the converse fails quite badly: e.g. $(mathbb{N}, <)$ is a reduct$_2$, but not a reduct$_1$, of $(mathbb{N},+)$.
The motivation for the terminology overload is that from a model-theoretic perspective, there is little difference between the "primitive" relations/functions/constants and the definable relations/functions/constants. For example, taking definitional expansions preserves: decidability, saturation, homogeneity, minimality and its various cousins, stability, ... So much of the time in model theory we lose nothing by conflating a given structure with its "full definitional expansion" (add all the $emptyset-$definable relations/functions/constants), and then the two notions of "reduct" essentially coincide.
But that said, it is still an overload of terminology and can definitely lead to confusion if one is not careful. That's unfortunate, but that's the situation.
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
answered Dec 28 '18 at 3:38
Noah SchweberNoah Schweber
125k10150288
125k10150288
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
add a comment |
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
$begingroup$
Thank you for your comprehensive answer Noah! So if consider $mathcal{M}^*= (M, f_1, dots, R_1, dots, c_1,dots)$ which is the full expansion of $mathcal{M}$ obtained by adding all $emptyset$-definable functions/relations/constants, then the notion of reduct$_1$ and reduct$_2$ coincide. Right?
$endgroup$
– Alice.H
Dec 28 '18 at 17:03
add a comment |
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Why do you believe the definition of "reduct" will be identical in similar (but very slightly different) settings?
$endgroup$
– Eric Towers
Dec 28 '18 at 2:31