Does this formalism adequately describe functions of one variable?












2












$begingroup$


Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.






functions elementary-set-theory logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    5 hours ago






  • 1




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    5 hours ago










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    4 hours ago
















2












$begingroup$


Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.






functions elementary-set-theory logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    5 hours ago






  • 1




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    5 hours ago










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    4 hours ago














2












2








2





$begingroup$


Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.






functions elementary-set-theory logic







share|cite|improve this question











$endgroup$




Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.



Can this statement be effectively formalized by




$forall a (ain X implies f(a) in Y)$




What logical aspects of functionality, if any, would not be captured by this statement.






functions elementary-set-theory logic




functions elementary-set-theory logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







Dan Christensen

















asked 5 hours ago









Dan ChristensenDan Christensen

8,63821835




8,63821835












  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    5 hours ago






  • 1




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    5 hours ago










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    4 hours ago


















  • $begingroup$
    Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
    $endgroup$
    – bounceback
    5 hours ago






  • 1




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    5 hours ago










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    5 hours ago










  • $begingroup$
    @Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
    $endgroup$
    – Dan Christensen
    4 hours ago
















$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
5 hours ago




$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
5 hours ago




1




1




$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
5 hours ago




$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
5 hours ago












$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
5 hours ago




$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
5 hours ago












$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
5 hours ago




$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
5 hours ago












$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
4 hours ago




$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
4 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
begin{align*}
(forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am familiar with the that particular formalism.Could you comment on the one I presented here?
    $endgroup$
    – Dan Christensen
    4 hours ago












  • $begingroup$
    I guess the heading wasn't very clear. I have changed it.
    $endgroup$
    – Dan Christensen
    4 hours ago



















3












$begingroup$

For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    3












    $begingroup$

    Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
    begin{align*}
    (forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
    end{align*}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am familiar with the that particular formalism.Could you comment on the one I presented here?
      $endgroup$
      – Dan Christensen
      4 hours ago












    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      4 hours ago
















    3












    $begingroup$

    Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
    begin{align*}
    (forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
    end{align*}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am familiar with the that particular formalism.Could you comment on the one I presented here?
      $endgroup$
      – Dan Christensen
      4 hours ago












    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      4 hours ago














    3












    3








    3





    $begingroup$

    Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
    begin{align*}
    (forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
    end{align*}






    share|cite|improve this answer









    $endgroup$



    Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
    begin{align*}
    (forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
    end{align*}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 5 hours ago









    APC89APC89

    2,443420




    2,443420












    • $begingroup$
      I am familiar with the that particular formalism.Could you comment on the one I presented here?
      $endgroup$
      – Dan Christensen
      4 hours ago












    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      4 hours ago


















    • $begingroup$
      I am familiar with the that particular formalism.Could you comment on the one I presented here?
      $endgroup$
      – Dan Christensen
      4 hours ago












    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      4 hours ago
















    $begingroup$
    I am familiar with the that particular formalism.Could you comment on the one I presented here?
    $endgroup$
    – Dan Christensen
    4 hours ago






    $begingroup$
    I am familiar with the that particular formalism.Could you comment on the one I presented here?
    $endgroup$
    – Dan Christensen
    4 hours ago














    $begingroup$
    I guess the heading wasn't very clear. I have changed it.
    $endgroup$
    – Dan Christensen
    4 hours ago




    $begingroup$
    I guess the heading wasn't very clear. I have changed it.
    $endgroup$
    – Dan Christensen
    4 hours ago











    3












    $begingroup$

    For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



    Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



    $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



    If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



    1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



      Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



      $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



      If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



      1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



        Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



        $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



        If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



        1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.






        share|cite|improve this answer











        $endgroup$



        For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.



        Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.



        $f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.



        If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)



        1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 3 hours ago

























        answered 3 hours ago









        Derek ElkinsDerek Elkins

        17k11437




        17k11437






























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            Fluorita

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            Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
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            Hulsita

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            Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
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            Península de Txukotka

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            Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
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