Restriction of Mapping is Subclass of Cartesian Product

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