Well-Ordered Induction

Theorem
Let $\left({A,\prec}\right)$ be a strict well-ordering.

For all $x \in A$, let the $\prec$-initial segment of $x$ be a small class.

Let $B$ be a class such that $B \subseteq A$.

Let:
 * $(1): \quad \forall x \in A: \left({ \left({ A \cap \prec^{-1} \left({ x }\right) }\right) \subseteq B \implies x \in B }\right)$

Then:


 * $A = B$

That is, if a property passes from the initial segment of $x$ to $x$, then this property is true for all $x \in A$.

Proof
that $A \not \subseteq B$.

Then:
 * $A \setminus B \ne 0$.

By Proper Well-Ordering Determines Smallest Elements, $A \setminus B$ must have some $\prec$-minimal element.

Thus:
 * $\displaystyle \exists x \in \left({ A \setminus B }\right): \left({ A \setminus B }\right) \cap \prec^{-1} \left({ x }\right) = \varnothing$

implies that:
 * $A \cap \prec^{-1} \left({ x }\right) \subseteq B$

Hence this fulfils the hypothesis for $(1)$.

We have that $x \in A$.

so by $(1)$:
 * $x \in B$

But this contradicts the fact that $x \in \left({A \setminus B}\right)$.

Thus by Proof by Contradiction:
 * $A \subseteq B$.

It follows by definition of set equality that:
 * $A = B$

Also see

 * Well-Founded Induction shows that it is possible to weaken the hypotheses in order to drop the requirements that $\prec$ be well-ordering, replacing it with the requirement that $\prec$ be simply foundational (hence, the name well-founded induction) and to drop the requirement that the initial segments be sets (they may also be proper classes).


 * It is important to note that such an approach involves the use of the axiom of foundation.