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 $mathbbR$ and $L_mathbbR$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]

Dominant seventh chord in the major scale contains diminished triad of the seventh?

Is there a "higher Segal conjecture"?

How do I keep my slimes from escaping their pens?

How do I mention the quality of my school without bragging

Using et al. for a last / senior author rather than for a first author

ListPlot join points by nearest neighbor rather than order

Why are there no cargo aircraft with "flying wing" design?

Do I really need recursive chmod to restrict access to a folder?

How widely used is the term Treppenwitz? Is it something that most Germans know?

Problem drawing boxes with arrows in tikZ

Is it true that "carbohydrates are of no use for the basal metabolic need"?

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

What is the musical term for a note that continously plays through a melody?

What's the purpose of writing one's academic bio in 3rd person?

Were Kohanim forbidden from serving in King David's army?

Determinant is linear as a function of each of the rows of the matrix.

Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?

How can I fade player character when he goes inside or outside of the area?

What is this single-engine low-wing propeller plane?

What does '1 unit of lemon juice' mean in a grandma's drink recipe?

Do you forfeit tax refunds/credits if you aren't required to and don't file by April 15?

How to deal with a team lead who never gives me credit?

List *all* the tuples!

Why constant symbols in a language?



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 $mathbbR$ and $L_mathbbR$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]










2












$begingroup$


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 $mathfrakS_n$ over the set $mathbbZ$, 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 $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ 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 $mathfrakS_n$ as L-structure and $mathfrakS_n$ 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.










share|cite|improve this question









$endgroup$











  • $begingroup$
    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
    6 hours ago















2












$begingroup$


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 $mathfrakS_n$ over the set $mathbbZ$, 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 $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ 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 $mathfrakS_n$ as L-structure and $mathfrakS_n$ 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.










share|cite|improve this question









$endgroup$











  • $begingroup$
    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
    6 hours ago













2












2








2





$begingroup$


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 $mathfrakS_n$ over the set $mathbbZ$, 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 $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ 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 $mathfrakS_n$ as L-structure and $mathfrakS_n$ 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.










share|cite|improve this question









$endgroup$




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 $mathfrakS_n$ over the set $mathbbZ$, 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 $mathfrakS_0$ we would understand $c$ as $0$.
Since there are no relation- or functionsymbols, we just have the set $mathbbZ$ and could note them as



$dotso, -1, c, 1, dotso$



If we take the usual function $+$ and add it $L=c,+$ now $mathfrakS_0$ 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 $mathfrakS_n$ as L-structure and $mathfrakS_n$ 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 6 hours ago









CornmanCornman

3,75321233




3,75321233











  • $begingroup$
    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
    6 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
    6 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
6 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
6 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

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, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ 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 "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
    6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    6 hours ago



















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189303%2fwhy-constant-symbols-in-a-language%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ 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 "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
      6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      6 hours ago
















    5












    $begingroup$

    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, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ 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 "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
      6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      6 hours ago














    5












    5








    5





    $begingroup$

    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, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ 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 "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.






    share|cite|improve this answer









    $endgroup$



    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, "$mathbbZ$ as a group" and "$mathbbZ$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : mathbbZ times mathbbZ to mathbbZ$, which must be preserved by group homomorphisms.



    In your example, a homomorphism of $L$-structures $f : mathfrakS_n to mathfrakS_m$ 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 "$mathfrakS_n$ as an $L$-structure" and "$mathfrakS_n$ as an $L_varnothing$-structure" have the same underlying set, they are not the same object.



    Fun fact: the assignment from "$mathfrakS_n$ as an $L$-structure" to "$mathfrakS_n$ as an $L_varnothing$-structure" is an example of a forgetful functor.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 6 hours ago









    Clive NewsteadClive Newstead

    52.2k474137




    52.2k474137











    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
      6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      6 hours ago

















    • $begingroup$
      Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
      6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      6 hours ago
















    $begingroup$
    Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
    6 hours ago




    $begingroup$
    Now I wonder, if structures $mathfrakS_n, mathfrakS_m$ 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
    6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    6 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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    6 hours ago












    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        Mark KamsmaMark Kamsma

        3918




        3918



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189303%2fwhy-constant-symbols-in-a-language%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Möglingen Índice Localización Historia Demografía Referencias Enlaces externos Menú de navegación48°53′18″N 9°07′45″E / 48.888333333333, 9.129166666666748°53′18″N 9°07′45″E / 48.888333333333, 9.1291666666667Sitio web oficial Mapa de Möglingen«Gemeinden in Deutschland nach Fläche, Bevölkerung und Postleitzahl am 30.09.2016»Möglingen

            Virtualbox - Configuration error: Querying “UUID” failed (VERR_CFGM_VALUE_NOT_FOUND)“VERR_SUPLIB_WORLD_WRITABLE” error when trying to installing OS in virtualboxVirtual Box Kernel errorFailed to open a seesion for the virtual machineFailed to open a session for the virtual machineUbuntu 14.04 LTS Virtualbox errorcan't use VM VirtualBoxusing virtualboxI can't run Linux-64 Bit on VirtualBoxUnable to insert the virtual optical disk (VBoxguestaddition) in virtual machine for ubuntu server in win 10VirtuaBox in Ubuntu 18.04 Issues with Win10.ISO Installation

            Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org