Restriction of Mapping is Subclass of Cartesian Product
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Theorem
Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $A$ be a class.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Let $f {\restriction} A$ denote the restriction of $f$ to $A$.
Then $f {\restriction} A$ is a subclass of the cartesian product of $A$ with its image:
- $f {\restriction} A \subseteq A \times f \sqbrk A$
Proof
Follows directly from:
- the definition of restriction
- the definition of 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