Image of Union of Nest of Mappings is Union of Class of Images

Theorem

Let $N$ be a nest of mappings.

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

Then:

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

where $\Img f$ denotes the image of $f$.

Proof

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

Let $y \in \Img {\bigcup N}$.

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:

$\exists f \subseteq \bigcup N: y \in \Img f$

That is:

$y \in \ds \bigcup_{f \mathop \in N} \Img f$

That is:

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

Let $y \in \ds \bigcup_{f \mathop \in N} \Img f$.

Then:

$\exists f \subseteq \bigcup N: y \in \Img f$

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:

$y \in \Img {\bigcup N}$

That is:

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

Hence by definition of set equality:

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

$\blacksquare$

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 $1.4$