Well-Founded Proper Relational Structure Determines Minimal Elements
Theorem
Let $A$ and $B$ be classes.
Let $\struct {A, \prec}$ be a proper relational structure.
Let $\prec$ be a strictly well-founded relation.
Suppose $B \subset A$ and $B \ne \O$.
Then $B$ has a strictly minimal element under $\prec$.
Proof
$B$ is not empty.
So $B$ has at least one element $x$.
By Singleton of Element is Subset:
- $\set x \subseteq B$
By Relational Closure Exists for Set-Like Relation:
- $\set x$ has a $\prec$-relational closure.
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This $\prec$-relational closure shall be denoted $y$.
- $y \cap B \subseteq A$
As $x \in y$ and $x \in B$:
- $y \cap B \ne \O$
By the definition of strictly well-founded relation:
- $(1): \quad \exists x \in \paren {y \cap B}: \forall w \in \paren {y \cap B}: w \nprec x$
Aiming for a contradiction, suppose that $w \in B$ such that $w \prec x$.
Since $x \in y$, it follows that $w \in y$, so $w \in \paren {B \cap y}$ by the definition of intersection.
This contradicts $(1)$.
Therefore:
- $w \nprec x$
and so $B$ has a strictly minimal element under $\prec$.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 9.4$