Subclass of Well-Ordered Class is Well-Ordered

Theorem

Let $V$ be a basic universe.

Let $\RR \subseteq V \times V$ be a relation.

Let $A$ be a subclass of $V$ which is well-ordered under $\RR$.

Let $B$ be a subclass of $A$.

Then $B$ is also well-ordered under $\RR$.

Proof

First suppose $B$ is the empty class.

From Empty Class is Subclass of All Classes, $B$ is a subclass of $A$.

Then by Empty Class is Well-Ordered, $\O$ is well-ordered under $\RR$.

Otherwise, let $X$ be an arbitrary non-empty class subclass of $B$.

By Subclass of Subclass is Subclass, $X$ is a subclass of $A$.

Hence by definition of well-ordered class, $X$ has a smallest element under $\RR$.

As $X$ is arbitrary, it follows that every non-empty class subclass of $B$ has a smallest element under $\RR$.

Hence the result by definition of well-ordered class.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.1$