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]

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

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

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

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

Right-skewed distribution with mean equals to mode?

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

Gastric acid as a weapon

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

Why did the IBM 650 use bi-quinary?

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

Is the address of a local variable a constexpr?

Center align columns in table ignoring minus signs?

Why is "Captain Marvel" translated as male in Portugal?

What LEGO pieces have "real-world" functionality?

Doubts about chords

What are the pros and cons of Aerospike nosecones?

Did Kevin spill real chili?

Does surprise arrest existing movement?

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

Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?

Bonus calculation: Am I making a mountain out of a molehill?

Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?

Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?

Single word antonym of "flightless"



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
    3 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
    3 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
    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




$begingroup$
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










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

            Magento 2 duplicate PHPSESSID cookie when using session_start() in custom php scriptMagento 2: User cant logged in into to account page, no error showing!Magento duplicate on subdomainGrabbing storeview from cookie (after using language selector)How do I run php custom script on magento2Magento 2: Include PHP script in headerSession lock after using Cm_RedisSessionscript php to update stockMagento set cookie popupMagento 2 session id cookie - where to find it?How to import Configurable product from csv with custom attributes using php scriptMagento 2 run custom PHP script

            Can not update quote_id field of “quote_item” table magento 2Magento 2.1 - We can't remove the item. (Shopping Cart doesnt allow us to remove items before becomes empty)Add value for custom quote item attribute using REST apiREST API endpoint v1/carts/cartId/items always returns error messageCorrect way to save entries to databaseHow to remove all associated quote objects of a customer completelyMagento 2 - Save value from custom input field to quote_itemGet quote_item data using quote id and product id filter in Magento 2How to set additional data to quote_item table from controller in Magento 2?What is the purpose of additional_data column in quote_item table in magento2Set Custom Price to Quote item magento2 from controller

            How to solve knockout JS error in Magento 2 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?(Magento2) knockout.js:3012 Uncaught ReferenceError: Unable to process bindingUnable to process binding Knockout.js magento 2Cannot read property `scopeLabel` of undefined on Product Detail PageCan't get Customer Data on frontend in Magento 2Magento2 Order Summary - unable to process bindingKO templates are not loading in Magento 2.1 applicationgetting knockout js error magento 2Product grid not load -— Unable to process binding Knockout.js magento 2Product form not loaded in magento2Uncaught ReferenceError: Unable to process binding “if: function()return (isShowLegend()) ” magento 2