# 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 $f(x)=\bigcap W$.

$$x\notin \mbox{def } f \Longleftrightarrow f(x)=\mathcal U,$$

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 $\mathcal U$.

• I would certainly expect $x\in\operatorname{def}f\implies f(x)\in\mathcal U$ to hold, which seems hardly compatible with the iven claim. – Hagen von Eitzen Jan 21 '18 at 16:43
• The original claim does not look right at all. The $\Rightarrow$ direction might be true depending on how exactly you have defined the notation $f(x)$ -- but the $\Leftarrow$ direction is nonsense: Just because $f(x)$ is a set does not imply that $f$ is not defined at $x$, with any sensible definitions. – hmakholm left over Monica Jan 21 '18 at 16:44
• This is the second part of the theorem. And his proof is the same. Now $W\neq\emptyset$ and by a previous theorem, $\bigcap W$ is a set (the theoreom is a corollary of subset axiom and the fact that if $y\in x$, then $\bigcap x\subset y$. – Dog_69 Jan 21 '18 at 16:44
• Also, I would expect $f(x)=\bigcap W$ to hold under normal circumstances (but also $f(x)=\bigcup W$) – Hagen von Eitzen Jan 21 '18 at 16:46
• @MauroALLEGRANZA: That is a great answer. You can consider write it as an answer and I'll mark it. Thanks a lot. Moreover, that comments also helps me with my another question math.stackexchange.com/questions/2618981/…. – Dog_69 Jan 24 '18 at 15:42

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.