Union of Nest of Mappings is Mapping/Proof

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


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

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.