Sandwich Principle/Proof 1

Proof
We are given that:


 * for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.

Let $x, y \in A$ such that:
 * $x \subseteq y \subseteq \map g x$

both $x \subset y$ and $y \subset \map g x$

From $x \subset y$, it follows by definition of proper subset that:
 * $\exists a \in y: a \notin x$

and so it is not the case that $y \subseteq x$.

From $y \subset \map g x$, it follows by definition of proper subset that:
 * $\exists b \in \map g x: b \notin y$

and so it is not the case that $\map g x \subseteq y$.

Hence neither $y \subseteq x$ nor $\map g x \subseteq y$.

This contradicts the property of $g$.

Hence by Proof by Contradiction it follows that $y \subset \map g x$ and $x \subset y$ cannot both be true.

Therefore either $x = y$ or $y = \map g x$.

Hence the result.