Image of Union of Nest of Mappings is Union of Class of Images/Proof
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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: Exercise $1.1$