6
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ They're compatible as long as $x \in \operatorname{dom} f$ $\endgroup$
    – wlad
    Jan 24, 2018 at 11:31
  • $\begingroup$ They also assume that the $y$s are sets $\endgroup$
    – wlad
    Jan 24, 2018 at 11:31
  • $\begingroup$ Is $\text{dom}$ short for domain? I have never seen that notation before. $\endgroup$
    – Mr Pie
    Jan 24, 2018 at 12:04
  • $\begingroup$ @user477343: Yes. That notation is standard all across mathematics. $\endgroup$
    – Asaf Karagila
    Jan 24, 2018 at 12:17
  • $\begingroup$ @AsafKaragila thank you :) $\endgroup$
    – Mr Pie
    Jan 24, 2018 at 16:40

1 Answer 1

8
$\begingroup$

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 $y$.

$\endgroup$
20
  • 5
    $\begingroup$ I'm not a set! I'm a free number! $\endgroup$ Jan 24, 2018 at 11:35
  • 3
    $\begingroup$ You are number $\{0,1,2,3,4,5\}$. $\endgroup$
    – Asaf Karagila
    Jan 24, 2018 at 11:36
  • 2
    $\begingroup$ I don't see where you get the fact that he is $6$? Any hints? $\endgroup$ Jan 24, 2018 at 11:38
  • 1
    $\begingroup$ @WeijunZhou: Then you didn't get the joke... :) $\endgroup$
    – Asaf Karagila
    Jan 24, 2018 at 11:39
  • $\begingroup$ @WeijunZhou Hint $\endgroup$ Jan 24, 2018 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.