Why constant symbols in a language? Announcing the arrival of Valued Associate #679: Cesar...
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Why constant symbols in a language?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why does ${phi,(phiRightarrowpsi)}$ not semantically entail $psi$ if $phi$ has a free variable and $psi$ doesn't?satisfiability in a structure implies satisfiability in a substructure?Eventually constant variable assignmentsTerm models in group theoryLanguage structure of $mathbb{R}$ and $L_{mathbb{R}}$Structure/Model for first order languageExtending a language by adding a constant symbolDo we really need constant symbols in first-order theories?Logic: one vs many structures for a given languageIntuition behind a structure of a language in mathematical logic [long read but simple]
$begingroup$
What is the point of constant symbols in a language?
For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?
I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.
Is there something else, which makes constants 'special'?
Other example: Suppose we have the language $L={c}$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrak{S}_n$ over the set $mathbb{Z}$, where $c$ gets interpreted by $n$.
Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?
For $mathfrak{S}_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbb{Z}$ and could note them as
${dotso, -1, c, 1, dotso}$
If we take the usual function $+$ and add it $L={c,+}$ now $mathfrak{S}_0$ has the property, that $c+c=c$ for example.
I hope you understand what I am asking for.
I think it boils down to:
Is there a difference between the structure $mathfrak{S}_n$ as L-structure and $mathfrak{S}_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).
But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)
Thanks in advance.
logic first-order-logic
$endgroup$
add a comment |
$begingroup$
What is the point of constant symbols in a language?
For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?
I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.
Is there something else, which makes constants 'special'?
Other example: Suppose we have the language $L={c}$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrak{S}_n$ over the set $mathbb{Z}$, where $c$ gets interpreted by $n$.
Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?
For $mathfrak{S}_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbb{Z}$ and could note them as
${dotso, -1, c, 1, dotso}$
If we take the usual function $+$ and add it $L={c,+}$ now $mathfrak{S}_0$ has the property, that $c+c=c$ for example.
I hope you understand what I am asking for.
I think it boils down to:
Is there a difference between the structure $mathfrak{S}_n$ as L-structure and $mathfrak{S}_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).
But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)
Thanks in advance.
logic first-order-logic
$endgroup$
$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila♦
3 hours ago
add a comment |
$begingroup$
What is the point of constant symbols in a language?
For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?
I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.
Is there something else, which makes constants 'special'?
Other example: Suppose we have the language $L={c}$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrak{S}_n$ over the set $mathbb{Z}$, where $c$ gets interpreted by $n$.
Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?
For $mathfrak{S}_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbb{Z}$ and could note them as
${dotso, -1, c, 1, dotso}$
If we take the usual function $+$ and add it $L={c,+}$ now $mathfrak{S}_0$ has the property, that $c+c=c$ for example.
I hope you understand what I am asking for.
I think it boils down to:
Is there a difference between the structure $mathfrak{S}_n$ as L-structure and $mathfrak{S}_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).
But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)
Thanks in advance.
logic first-order-logic
$endgroup$
What is the point of constant symbols in a language?
For example we take the language of rings $(0,1,+,-,cdot)$.
What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?
I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1cdot x = xcdot 1=x$.
Is there something else, which makes constants 'special'?
Other example: Suppose we have the language $L={c}$ where $c$ is a constant symbol.
Now we observe the L-structure $mathfrak{S}_n$ over the set $mathbb{Z}$, where $c$ gets interpreted by $n$.
Is there any difference, between $c$ and $n$?
Or are they just the same and you can view it as some sort of substitution?
For $mathfrak{S}_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbb{Z}$ and could note them as
${dotso, -1, c, 1, dotso}$
If we take the usual function $+$ and add it $L={c,+}$ now $mathfrak{S}_0$ has the property, that $c+c=c$ for example.
I hope you understand what I am asking for.
I think it boils down to:
Is there a difference between the structure $mathfrak{S}_n$ as L-structure and $mathfrak{S}_n$ as $L_emptyset$-structure, where $L_emptyset=emptyset$ (so does not contain a constant symbol).
But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)
Thanks in advance.
logic first-order-logic
logic first-order-logic
asked 4 hours ago
CornmanCornman
3,74821233
3,74821233
$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila♦
3 hours ago
add a comment |
$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila♦
3 hours ago
$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila♦
3 hours ago
$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila♦
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.
For example, "$mathbb{Z}$ as a group" and "$mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbb{Z} times mathbb{Z} to mathbb{Z}$, which must be preserved by group homomorphisms.
In your example, a homomorphism of $L$-structures $f : mathfrak{S}_n to mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{varnothing}$-structures would not.
So while "$mathfrak{S}_n$ as an $L$-structure" and "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" have the same underlying set, they are not the same object.
Fun fact: the assignment from "$mathfrak{S}_n$ as an $L$-structure" to "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" is an example of a forgetful functor.
$endgroup$
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
1
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
add a comment |
$begingroup$
Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.
For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.
For example, "$mathbb{Z}$ as a group" and "$mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbb{Z} times mathbb{Z} to mathbb{Z}$, which must be preserved by group homomorphisms.
In your example, a homomorphism of $L$-structures $f : mathfrak{S}_n to mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{varnothing}$-structures would not.
So while "$mathfrak{S}_n$ as an $L$-structure" and "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" have the same underlying set, they are not the same object.
Fun fact: the assignment from "$mathfrak{S}_n$ as an $L$-structure" to "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" is an example of a forgetful functor.
$endgroup$
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
1
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
add a comment |
$begingroup$
An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.
For example, "$mathbb{Z}$ as a group" and "$mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbb{Z} times mathbb{Z} to mathbb{Z}$, which must be preserved by group homomorphisms.
In your example, a homomorphism of $L$-structures $f : mathfrak{S}_n to mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{varnothing}$-structures would not.
So while "$mathfrak{S}_n$ as an $L$-structure" and "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" have the same underlying set, they are not the same object.
Fun fact: the assignment from "$mathfrak{S}_n$ as an $L$-structure" to "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" is an example of a forgetful functor.
$endgroup$
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
1
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
add a comment |
$begingroup$
An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.
For example, "$mathbb{Z}$ as a group" and "$mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbb{Z} times mathbb{Z} to mathbb{Z}$, which must be preserved by group homomorphisms.
In your example, a homomorphism of $L$-structures $f : mathfrak{S}_n to mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{varnothing}$-structures would not.
So while "$mathfrak{S}_n$ as an $L$-structure" and "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" have the same underlying set, they are not the same object.
Fun fact: the assignment from "$mathfrak{S}_n$ as an $L$-structure" to "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" is an example of a forgetful functor.
$endgroup$
An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.
For example, "$mathbb{Z}$ as a group" and "$mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbb{Z} times mathbb{Z} to mathbb{Z}$, which must be preserved by group homomorphisms.
In your example, a homomorphism of $L$-structures $f : mathfrak{S}_n to mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{varnothing}$-structures would not.
So while "$mathfrak{S}_n$ as an $L$-structure" and "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" have the same underlying set, they are not the same object.
Fun fact: the assignment from "$mathfrak{S}_n$ as an $L$-structure" to "$mathfrak{S}_n$ as an $L_{varnothing}$-structure" is an example of a forgetful functor.
answered 4 hours ago
Clive NewsteadClive Newstead
52.2k474137
52.2k474137
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
1
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
add a comment |
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
1
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
$begingroup$
Now I wonder, if structures $mathfrak{S}_n, mathfrak{S}_m$ can be isomorphic (for $nneq m$) when you take $L={c,<}$. Then it has to be $n<mLeftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $zneq m,n$ it is $f(z)=z$. But then you get $n<mLeftrightarrow f(n)<f(m)$ so $n<mLeftrightarrow m<n$?
$endgroup$
– Cornman
4 hours ago
1
1
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
$begingroup$
@Cornman: Right, so what that tells you is that if $m ne n$ then there is no isomorphism of ${ c, < }$-structures between $mathfrak{S}_n$ and $mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbb{Z}$ in both $mathfrak{S}_n$ and $mathfrak{S}_m$).
$endgroup$
– Clive Newstead
3 hours ago
add a comment |
$begingroup$
Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.
For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.
$endgroup$
add a comment |
$begingroup$
Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.
For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.
$endgroup$
add a comment |
$begingroup$
Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.
For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.
$endgroup$
Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.
For example, if we consider Peano Arithmetic then obviously $mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > bar n$ for all $n in mathbb N$ (where $bar n$ stands for 1 added $n$ times: $1 + 1 + ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.
answered 3 hours ago
Mark KamsmaMark Kamsma
3818
3818
add a comment |
add a comment |
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RANn,iQfzSopy,SfusBmuEhYpOZumXm3,pWD 3kEN16n4fSwN
$begingroup$
In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element.
$endgroup$
– Asaf Karagila♦
3 hours ago