Well-Ordered Induction

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

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

Let $B \subseteq A$.

Let $\forall x \in A: ( ( A \cap A_x ) \subseteq B \implies x \in B )$. (1)

Then, $A = B$.

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

Proof
Assume, to the contrary, that $A \not \subseteq B$. Then $A \setminus B \not = 0$ and by Well-Founded Relation Determines Minimal Elements/Special Case, $A \setminus B$ must have some $\prec$-minimal element.


 * $\displaystyle \exists x \in ( A \setminus B ): ( A \setminus B ) \cap A_x = \varnothing$ implies that $A \cap A_x \subseteq B$. Notice that this satisfies the hypothesis for (1).

$x \in A$, so by (1), $x \in B$. But this contradicts the fact that $x \in (A \setminus B)$. Therefore, we are forced to conclude that $(A \setminus B) = \varnothing$ and $A \subseteq B$. Therefore, $A = B$.

Remark
With extra work, 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 well-founded (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 also important to note that such an approach involves the use of the axiom of foundation.