Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable/Mistake

Source Work

 * Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:
 * $\S 1$ Introduction to well ordering:
 * Proposition $1.2$: Proof
 * Proposition $1.2$: Proof

Mistake

 * Finally, suppose $C$ is any non-empty subclass of $A$. Let $C'$ be the class of all elements $x'$ such that $x \in C$. Then $C'$ contains a least element ( $R$), and this element is $b'$ for some $b \in C$. Then for any $x \in C$, $x' \in C'$, and so $x' R b'$, and therefore $x \le b$. Thus $b$ is the least element of $C$ (with respect to $\le$).

Correction
As $b'$ is the least element of $C'$, it follows that for arbitrary $x' \in C'$ that $b' R x'$, not $x' R b'$.

Hence similarly $b \le x$, not $x \le b$.

However, the conclusion is correct.

Also see

 * Class Equivalent to Subclass of Well-Orderable Class is Well-Orderable: 's rendition of this result