Image of Bijection on Set is Set with Axiom of Choice implies Axiom of Replacement

Theorem

Let the Axiom of Choice be accepted.

Let it be accepted as axiomatic that:

every class which can be put into one-to-one correspondence with a set.

Then the Axiom of Replacement holds:

For every mapping $f$ and set $x$ in the domain of $f$, the image $f \sqbrk x$ is a set.

Symbolically:

$\forall Y: \map {\text{Fnc}} Y \implies \forall x: \exists y: \forall u: u \in y \iff \exists v: \tuple {v, u} \in Y \land v \in x$

where:

$\map {\text{Fnc}} X := \forall x, y, z: \tuple {x, y} \in X \land \tuple {x, z} \in X \implies y = z$

and the notation $\tuple {\cdot, \cdot}$ is understood to represent Kuratowski's formalization of ordered pairs.

Proof

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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: $\S 3$ The axiom of substitution: Exercise $3.2$