Relation is a function from domain to power set of range

Question

Let $E$ and $F$ be sets. Then $\tau$ can be considered a function from $E$ to $P(F)$ by setting, for each $x \in E$,

$\tau(x) = \{y \in F: (x, y) \in \tau\}$ .

This is a claim from a text, but it doesn't make sense to me. How can $\tau$ be a mapping into $P(F)$ when the elements of its range are members of $F$? If the claim is true, shouldn't $\tau(x)$ be a subset of $F$ rather than an element of $F$?

Answer 1

Consider the "less than" relation on the integers. Let $x$ be the number $7$. Then the author is saying that we can write $$ \tau(7) = \{ y \in \mathbb Z \mid 7 < y \} = \{8, 9, 10, \ldots\} $$

A typical element in the range of $\tau$ is the set above, which is an element of the power set of $\mathbb Z$.

Does that make sense?

Answer 2

This is true in general.

The point is not that $\tau$ is a function into the powerset of its range — it isn't, it can't be, $\tau \subseteq E\times F$ — but that it induces such a function $\tau'$ in a natural, canonical way.

Note that the domain of $\tau'$ might not be all of $E$, but the domain of $\tau'$ will be all of $E$: $x\notin dom(\tau) = \{x\in E\mid \exists y\in F)(x,y)\in \tau\}$, iff $\tau'(x) = \emptyset$. So $\tau'$ is always a total function on $E$.

You can recover $\tau$ from $\tau'$: $$ \tau = \{(x,y)\mid y\in \tau'(x)\}. $$ Moreover, every relation from $E$ to $F$ is obtained from a function $E\to \mathcal{P}(F)$ in this way.