Union of Nest of Mappings is Mapping/Proof

Theorem

Let $N$ be a nest of mappings.

Then:

$\bigcup N$ is a mapping

where $\bigcup N$ denotes the union of $N$.

Proof

By definition of union, the elements of $\bigcup N$ are elements of elements of $N$.

That is, the elements of $\bigcup N$ are ordered pairs of sets.

Let $x \in \bigcup N$.

Then:

$\exists f \subseteq \bigcup N: \exists y: \tuple {x, y} \in f$

Aiming for a contradiction, suppose:

$\exists y_1, y_2: \tuple {x, y_1} \in \bigcup N \land \tuple {x, y_2} \in \bigcup N$

such that:

$y_1 \ne y_2$

Then:

$\exists f \subseteq \bigcup N: \tuple {x, y_1} \in f$

and:

$\exists g \subseteq \bigcup N: \tuple {x, y_2} \in g$

But because $N$ is a nest of mappings, either:

$f \subseteq g$

or:

$g \subseteq f$

That means either:

$\tuple {x, y_1} \in g$

or:

$\tuple {x, y_2} \in f$

Hence either $f$ or $g$ has both $\tuple {x, y_1}$ and $\tuple {x, y_2}$ in it.

That is, either $f$ or $g$ is not a mapping.

From this contradiction it follows that if:

$\exists y_1, y_2: \tuple {x, y_1} \in \bigcup N \land \tuple {x, y_2} \in \bigcup N$

then it has to be the case that:

$y_1 = y_2$

It follows that $\bigcup N$ is a mapping.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries: Exercise $1.1$