Set Complements flip subset ordering

up:: Set Complement

Let $A\subseteq B\subseteq X$, with $X$ the base space.

Then we have that $x\in A⟹x\in B$.

By contrapositive, we have that $x\notin B⟹x\notin A$¹. Thus, $B^c\subseteq A^c$.

Footnotes

1. And pressuposing always that $x\in X$. ↩