Proper Well-Ordering determines Smallest Elements

Theorem
Let $\prec$ well-order $S$.

Let $B$ be a nonempty subset of $S$.

For all $x \in S$, let every $\prec$-initial segment of $x$ be a set. We shall denote the initial segment of $x$

Then $B$ has an $\prec$-minimal element.

That is, there is an element $a \in B$ such that $B$ and the $\lt$-initial segment of $a$ are disjoint.

Proof
Because $B \ne \varnothing$ it follows that $\exists x \in B$.

If $B \cap S_x = \varnothing$, then then $x$ is a minimal element and the statement holds trivially.

Let $B \cap S_x \ne \varnothing$.


 * By the hypothesis, $B \cap S_x$ is a set and is a subset of $B$.
 * Moreover, $(B,\lt)$ is foundational by Foundational Relation Subset.

By the definition of a foundational relation, we can infer that


 * $\displaystyle \exists a \in B \cap S_x: ( B \cap S_x \cap S_a ) = \varnothing$

But $a \in S_x$, so $a \prec x$. This means that $S_a \subset S_x$ so we may remove $S_x$ from the intersection.


 * $\displaystyle \exists a \in B: ( B \cap S_a ) = \varnothing$

Source

 * : $\S 6.26$