Image of Set under Mapping is Set

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $A$ be a class.

Let $\mathrm U$ denote the universal class.

Let $f: A \to \mathrm U$ be a class mapping.

Let $S$ be a subset of $A$.



Then the image $f \sqbrk S$ is also a set.


If $A$ is a set, then this result is known as the Axiom of Replacement in ZF.


Proof

NotZFC.jpg

This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.

If you see any proofs that link to this page, please insert this template at the top.

If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.


Aiming for contradiction, suppose that $f \sqbrk S$ is not a set.

Then $f \sqbrk S$ must be proper.


By Restriction of Mapping to Image is Surjection, the restriction $f \restriction_{S \times f \sqbrk S}$ is a surjection.

But this contradicts Surjection from Class to Proper Class.

Thus by contradiction, $f \sqbrk S$ is a set.


Hence the result.

$\blacksquare$