Non-Empty Class has Element of Least Rank

From ProofWiki
Jump to navigation Jump to search


Let $C$ be a class.

Let $C \ne \varnothing$.

Then $C$ has an element of least rank.

That is:

$\exists x \in C: \forall y \in C: \operatorname{rank} \left({x}\right) \le \operatorname {rank}\left({y}\right)$

where $\operatorname{rank}\left({x}\right)$ is the rank of $x$.



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.

By Set has Rank, each element of $C$ has a rank.

Let $R$ be the class of ranks of elements of $C$.

$R$ is non-empty because $C$ is non-empty.

Since any non-empty class of ordinals has a least element, $R$ has a least element, $q$.

By the definition of $R$:

$\exists x \in C: \operatorname{rank}\left({x}\right) = q$

Then $x$ is an element of $C$ of least rank.