Sandwich Principle/Proof 2

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:
 * $x \subseteq y \subseteq \map g x$

Then either we have:
 * $\map g x \subseteq y$ and $y \subseteq \map g x$

in which case, by definition of set equality:
 * $y = \map g x$

or we have that:
 * $x \subseteq y$ and $y \subseteq x$

in which case, by definition of set equality:
 * $x = y$

Thus either $y = \map g x$ or $x = y$ and the result follows.