Domain of Union of Nest of Mappings is Union of Class of Domains/Proof

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Theorem

Let $N$ be a nest of mappings.

Let $\bigcup N$ denote the union of $N$.

Then:

$\Dom {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Dom f$

where $\Dom f$ denotes the domain of $f$.


Proof

From Union of Nest of Mappings is Mapping we have that $\bigcup N$ is a mapping.


Let $x \in \Dom {\bigcup N}$.

Then by definition of mapping:

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

Then by definition of union of class:

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

Hence:

$\exists f \subseteq \bigcup N: x \in \Dom f$

That is:

$x \in \ds \bigcup_{f \mathop \in N} \Dom f$

That is:

$\Dom {\bigcup N} \subseteq \ds \bigcup_{f \mathop \in N} \Dom f$


Let $x \in \ds \bigcup_{f \mathop \in N} \Dom f$.

Then:

$\exists f \subseteq \bigcup N: x \in \Dom f$

Then by definition of mapping:

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

Thus by definition of union of class:

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

It follows that:

$x \in \Dom {\bigcup N}$

That is:

$\ds \bigcup_{f \mathop \in N} \Dom f \subseteq \Dom {\bigcup N}$


Hence by definition of set equality:

$\Dom {\bigcup N} = \ds \bigcup_{f \mathop \in N} \Dom f$

$\blacksquare$


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