Non-Empty Class has Element of Least Rank

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Theorem

Let $C$ be a class.

Let $C \ne \O$.

Then $C$ has an element of least rank.

That is:

$\exists x \in C: \forall y \in C: \map {\operatorname {rank} } x \le \map {\operatorname {rank} } y$

where $\map {\operatorname {rank} } x$ is the rank of $x$.

Proof

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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.

This article, or a section of it, needs explaining. In particular: This should follow from arbitrary intersections of ordinals being ordinals. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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

By the definition of $R$:

$\exists x \in C: \map {\operatorname {rank} } x = q$

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

$\blacksquare$