In my book, a function $f$ is defined as a binary relation such that if $(x,y),(x,z)\in f$ then $y=z$. Moreover, as it is usual, author denotes $(x,y)\in f$ by

$$ y=f(x) . \tag{1} $$

So, from this notation, I understand $f(x)$ as the second component of the ordered pair $(x,y)\in f$, i.e $(x,f(x))$. Nevertheless, below, the author says that $f(x)=\bigcap\{y:(x,y)\in f\}$. But I think this notation is different from (1), since

$$ \bigcap\{y:(x,y)\in f\} = \{z: \forall y:(x,y)\in f \Longrightarrow z\in y\} , $$

I mean, $\bigcap\{y:(x,y)\in f\}$ is a set of elements of the class $y$, not such a $y$.

And, moreover, some other authors, as Herbert Enderton, define also $f(x)$ as the class

$$ f(x)=\bigcup\{y:(x,y)\in f\} . $$

How can be all these notations/definitions compatibles?

  • 1
    $\begingroup$ They're compatible as long as $x \in \operatorname{dom} f$ $\endgroup$
    – ogogmad
    Jan 24 '18 at 11:31
  • $\begingroup$ They also assume that the $y$s are sets $\endgroup$
    – ogogmad
    Jan 24 '18 at 11:31
  • $\begingroup$ Is $\text{dom}$ short for domain? I have never seen that notation before. $\endgroup$
    – Mr Pie
    Jan 24 '18 at 12:04
  • $\begingroup$ @user477343: Yes. That notation is standard all across mathematics. $\endgroup$
    – Asaf Karagila
    Jan 24 '18 at 12:17
  • $\begingroup$ @AsafKaragila thank you :) $\endgroup$
    – Mr Pie
    Jan 24 '18 at 16:40

If $A$ is a singleton, $\{a\}$ then $\bigcap A=\bigcup A=a$. Since $f$ is a function, the set $\{y : (x,y)\in f\}$ is a singleton, for a fixed $x$, so the result follows.

The key point to remember is that everything is a set, including $x$ and $y$.

  • 5
    $\begingroup$ I'm not a set! I'm a free number! $\endgroup$ Jan 24 '18 at 11:35
  • 3
    $\begingroup$ You are number $\{0,1,2,3,4,5\}$. $\endgroup$
    – Asaf Karagila
    Jan 24 '18 at 11:36
  • 2
    $\begingroup$ I don't see where you get the fact that he is $6$? Any hints? $\endgroup$ Jan 24 '18 at 11:38
  • 1
    $\begingroup$ @WeijunZhou: Then you didn't get the joke... :) $\endgroup$
    – Asaf Karagila
    Jan 24 '18 at 11:39
  • $\begingroup$ @WeijunZhou Hint $\endgroup$ Jan 24 '18 at 11:40

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