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 $B \subseteq A$.

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

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
Assume, to the contrary, that $A \not \subseteq B$.

Then $A \setminus B \not = 0$.

By Well-Founded Relation Determines Minimal Elements/Special Case, $A \setminus B$ must have some $\prec$-minimal element.


 * $\displaystyle \exists x \in \left({ A \setminus B }\right): \left({ A \setminus B }\right) \cap \prec^{-1} \left({ x }\right) = \varnothing$ implies that $A \cap \prec^{-1} \left({ x }\right) \subseteq B$. Notice that this satisfies the hypothesis for (1).

$x \in A$, so by \left({1}\right), $x \in B$.

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

Therefore, we are forced to conclude that $\left({A \setminus B}\right) = \varnothing$ and $A \subseteq B$.

Therefore, $A = B$.

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.