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).