Image of $f(x)$ when $x$ is not in the domain of $f$. Joaquin Olivert's book Estructuras de álgebra multilineal (multilinear algebra sructures) states the following (p.38 thm. 5.9):
Let $f$ be a function (an special type of binary relation) and let $\mbox{def }f$ be the class the definition domain of $f$. Then:
$$
x\notin \mbox{def }f \Longrightarrow f(x) = \mathcal U
$$
and
$$
x\in\mbox{def } f \Longrightarrow f(x)\in \mathcal U .
$$
(here $\mathcal U$ denotes the class of all sets).
I have tried to prove it, but I have obtained something different (very different in fact). My attempt:
Let
$$
W=\{y : (x,y)\in f\}.
$$
By hypothesis $W=\emptyset$. Now, by definition, $f(x)$ is the second component of the class $W$, so:
$$
f(x)= \left(\bigcap\bigcap W\right) \cup \left(\left( \bigcup\bigcup W\right)\dashv \bigcup\bigcap W\right) ,
$$
where $A\dashv B$ denotes the complement of $B$ respect to $X$:
$$
A\dashv B= A\cap\{s:s\notin B\}.
$$
Computing:
$$
\bigcap\bigcap\emptyset = \bigcap \mathcal U = \emptyset.
$$
and
$$
\bigcup\bigcup \emptyset = \emptyset
$$
Since the complement involves the intersection with the emptyset, it turns out to be the emptyset again, so that
$$
f(x)=\emptyset.
$$
Any help? 
Thanks a lot.
EDIT:
In the author's proof, he considers that
$$
f(x)=\bigcap W=\bigcap \emptyset = \mathcal U ,
$$
but I don't understand why $f(x)=\bigcap W $.
Addendum. The theorem actually states
$$
x\notin \mbox{def } f \Longleftrightarrow f(x)=\mathcal U,
$$
but it seems to be a mistake. Only $\Rightarrow$ is true.
Addendum 2. There was a mistake in my statement. First result said
$$
x\notin \mbox{def } f \Longrightarrow f(x) \in \mathcal U
$$
but the truth is $f(x)$ EQUAL TO $\mathcal U$.
 A: Comments
We can see see H.Enderton, Elements of Set Theory, page 43: if $F$ is a function and $\langle x,y \rangle \in F$, there is a unique $y$ such that $\langle x,y \rangle \in F$. 
We call it "the value of $F$ at $x$" and we can denote it with $F(x)$. 
Because of this: $\{ y \mid \langle x,y \rangle \in F \} = \{ y \}$, and we know that $\bigcup \{ y \} = y$. 
Conclusion: $F(x)=\bigcup \{ y \mid \langle x,y \rangle \in F \}$.
For the easy part, see Enderton page 41: if $\langle x,y \rangle \in F$, then $x,y \in \bigcup \bigcup F$. Thus: $x$ and $y$ are sets, and thus $x \text { (and } y ) \in \mathcal U$.
Regarding the second part, the author uses: $W = \{ y \mid \langle x,y \rangle \in F \}$. 
But then, $W=\emptyset$ and $\bigcap W = \bigcap \emptyset$  (see Enderton, page 25) is the class $\text V$ of all sets (i.e. $\mathcal U$).
But the issue is: writing $F(x) = \mathcal U$ makes little sense, because again the "meaning" of $F(x)$ is "the unique $y$ such that $\langle x,y \rangle \in F$" and obviously $\langle x, \mathcal U \rangle \notin F$, otherwise (see above): $x, \mathcal U \in \bigcup \bigcup F$, that implies that $\mathcal U$ is a set.
