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]

Why does Python start at index -1 when indexing a list from the end?

Is 1 ppb equal to 1 μg/kg?

What LEGO pieces have "real-world" functionality?

ListPlot join points by nearest neighbor rather than order

Disable hyphenation for an entire paragraph

How to find all the available tools in macOS terminal?

What is the longest distance a 13th-level monk can jump while attacking on the same turn?

Why did the IBM 650 use bi-quinary?

Does surprise arrest existing movement?

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

What causes the vertical darker bands in my photo?

What do you call a phrase that's not an idiom yet?

Stars Make Stars

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

Check which numbers satisfy the condition [A*B*C = A! + B! + C!]

Did Kevin spill real chili?

How do I keep my slimes from escaping their pens?

Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?

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

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

How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?

What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?

What is the correct way to use the pinch test for dehydration?

Antler Helmet: Can it work?



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]










1












$begingroup$


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
    4 hours ago















1












$begingroup$


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
    4 hours ago













1












1








1





$begingroup$


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 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
    4 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
    4 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
4 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
4 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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
    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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    3 hours ago



















1












$begingroup$

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









    4












    $begingroup$

    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
      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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 hours ago
















    4












    $begingroup$

    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
      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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 hours ago














    4












    4








    4





    $begingroup$

    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 4 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
      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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 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
      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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
      $endgroup$
      – Clive Newstead
      3 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
    4 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
    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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    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 $mathfrakS_n$ and $mathfrakS_m$ (assuming that $<$ is interpreted as the usual order relation on $mathbbZ$ in both $mathfrakS_n$ and $mathfrakS_m$).
    $endgroup$
    – Clive Newstead
    3 hours ago












    1












    $begingroup$

    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









    $endgroup$

















      1












      $begingroup$

      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









      $endgroup$















        1












        1








        1





        $begingroup$

        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 3 hours ago









        Mark KamsmaMark Kamsma

        3818




        3818



























            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

            Dapidodigma demeter Subspecies | Notae | Tabula navigationisDapidodigmaAfrotropical Butterflies: Lycaenidae - Subtribe IolainaAmplifica

            Constantinus Vanšenkin Nexus externi | Tabula navigationisБольшая российская энциклопедияAmplifica

            Vas sanguineum Index Historia | Divisio | Constructio anatomica | Vasorum sanguineorum morbi (angiopathiae) | Notae | Nexus interniTabula navigationisAmplifica