Axiom of Replacement implies Image of Bijection on Set is Set
Jump to navigation
Jump to search
Theorem
Let the Axiom of Replacement (in the context of class theory) be accepted.
Let $A$ be a class which can be put into one-to-one correspondence with a set.
Then $A$ is a set.
Proof
Let $x$ be a set such that $A$ can be put into one-to-one correspondence with $x$.
Then by definition there exists a bijection from $x$ to $A$.
Let $f: x \to A$ be such a bijection.
Then by the Axiom of Replacement, $f \sqbrk x$ is a set.
But:
- $f \sqbrk x = A$
Hence the result.
$\blacksquare$
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: $\S 3$ The axiom of substitution: Exercise $3.2$