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
    5 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
    5 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
    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




    $begingroup$
    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
    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




$begingroup$
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
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
5 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

    votes






    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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            Knooppunt Holsloot

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            Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
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            Altaar (religie)

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            Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
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            Gregoriusmis

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            Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
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