Does this formalism adequately describe functions of one variable?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












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.










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
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @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
    Feb 20 at 1:45















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.










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
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @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
    Feb 20 at 1:45













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.










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 20 at 1:56







Dan Christensen

















asked Feb 20 at 1:15









Dan ChristensenDan Christensen

8,64321835




8,64321835











  • $begingroup$
    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
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @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
    Feb 20 at 1:45
















  • $begingroup$
    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
    Feb 20 at 1:24






  • 2




    $begingroup$
    Seems like you're missing the uniqueness of $f(a)$ given $a$.
    $endgroup$
    – Randall
    Feb 20 at 1:24










  • $begingroup$
    @Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
    $endgroup$
    – Dan Christensen
    Feb 20 at 1:36










  • $begingroup$
    Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
    $endgroup$
    – Randall
    Feb 20 at 1:38










  • $begingroup$
    @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
    Feb 20 at 1:45















$begingroup$
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
Feb 20 at 1:24




$begingroup$
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
Feb 20 at 1:24




2




2




$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24




$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24












$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36




$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36












$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38




$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38












$begingroup$
@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
Feb 20 at 1:45




$begingroup$
@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
Feb 20 at 1:45










3 Answers
3






active

oldest

votes


















5












$begingroup$

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












    $begingroup$

    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
    beginalign*
    (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
    endalign*






    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
      Feb 20 at 1:54











    • $begingroup$
      I guess the heading wasn't very clear. I have changed it.
      $endgroup$
      – Dan Christensen
      Feb 20 at 1:57


















    0












    $begingroup$

    If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
      $endgroup$
      – Dan Christensen
      Feb 21 at 15:43










    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    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%2f3119587%2fdoes-this-formalism-adequately-describe-functions-of-one-variable%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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











    $endgroup$

















      5












      $begingroup$

      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











      $endgroup$















        5












        5








        5





        $begingroup$

        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 Feb 20 at 2:59

























        answered Feb 20 at 2:46









        Derek ElkinsDerek Elkins

        17.3k11437




        17.3k11437





















            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
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*






            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
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57















            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
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*






            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
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57













            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
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*






            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
            beginalign*
            (forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
            endalign*







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 20 at 1:22









            APC89APC89

            2,341720




            2,341720











            • $begingroup$
              I am familiar with the that particular formalism.Could you comment on the one I presented here?
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57
















            • $begingroup$
              I am familiar with the that particular formalism.Could you comment on the one I presented here?
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:54











            • $begingroup$
              I guess the heading wasn't very clear. I have changed it.
              $endgroup$
              – Dan Christensen
              Feb 20 at 1:57















            $begingroup$
            I am familiar with the that particular formalism.Could you comment on the one I presented here?
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:54





            $begingroup$
            I am familiar with the that particular formalism.Could you comment on the one I presented here?
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:54













            $begingroup$
            I guess the heading wasn't very clear. I have changed it.
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:57




            $begingroup$
            I guess the heading wasn't very clear. I have changed it.
            $endgroup$
            – Dan Christensen
            Feb 20 at 1:57











            0












            $begingroup$

            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43















            0












            $begingroup$

            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43













            0












            0








            0





            $begingroup$

            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.






            share|cite|improve this answer









            $endgroup$



            If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 20 at 9:07









            Alberto TakaseAlberto Takase

            2,380719




            2,380719











            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43
















            • $begingroup$
              What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
              $endgroup$
              – Dan Christensen
              Feb 21 at 15:43















            $begingroup$
            What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
            $endgroup$
            – Dan Christensen
            Feb 21 at 15:43




            $begingroup$
            What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
            $endgroup$
            – Dan Christensen
            Feb 21 at 15:43

















            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%2f3119587%2fdoes-this-formalism-adequately-describe-functions-of-one-variable%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

            How to check contact read email or not when send email to Individual?

            Bahrain

            Postfix configuration issue with fips on centos 7; mailgun relay