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 5 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
    5 hours ago












  • $begingroup$
    I guess the heading wasn't very clear. I have changed it.
    $endgroup$
    – Dan Christensen
    5 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






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    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
      5 hours ago












    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      5 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
      5 hours ago












    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      5 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
      5 hours ago












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


















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












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
















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






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














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




    $begingroup$
    I guess the heading wasn't very clear. I have changed it.
    $endgroup$
    – Dan Christensen
    5 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 4 hours ago

























        answered 4 hours ago









        Derek ElkinsDerek Elkins

        17k11437




        17k11437






























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            الفوسفات في المغرب

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            قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
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            Four equal circles intersect: What is the area of the small shaded portion and its height

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            3 $begingroup$ In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. geometry circle share | cite | improve this question edited 2 hours ago Abdulhameed asked 2 hours ago ...
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            جامعة ليفربول

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            جامعة ليفربول University of Liverpool الأسماء السابقة كلية ليفربول الجامعية (1882-1884) جامعة فيكتوريا في ليفربول (1884-1903) معلومات التأسيس 1882 (منذ 137 سنة) الإنتماءات رابطة الجامعات الأوروبية مجموعة راسل جامعات القرميد الأحمر المنحة المالية 143.5 مليون £ النوع جامعة عامة الكليات 6 الموقع الجغرافي إحداثيات 53°24′22″N 2°58′01″W  /  53.406°N 2.967°W  / 53.406; -2.967   المدينة ليفربول المكان   إنجلترا البلد   المملكة المتحدة رقم الهاتف 7942000 151 (44+) إحصاءات الطلاب 17,405 (سنة 2008) عضوية مجموعة رسل [1] أورسيد [2]   طلاب الدراسات العليا 3,860 الموقع www.liverpool.ac.uk تعديل   جامعة ليفربول مبنى فكتوريا جامعة ليفربول (بالإنجليزية: The University of Liverpool )) إحدى الجامعات البريطانية العريقة، تقع في مدينة ليفربول على الساحل الشمالي الغربي من أنجلترا . وهي عضو في مجموعة راسل للجامعات وتعتبر من ضمن جامعات الطوب الأحمر الستة. محتويات ...
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