Union of Nest of Mappings is Mapping
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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: Proposition $1.4$