Set Complement inverts Subsets/Corollary

Corollary to Set Complement inverts Subsets

Let $S$ and $T$ be sets.

Then:

$S \subseteq \map \complement T \iff T \subseteq \map \complement S$

where:

$S \subseteq \map \complement T$ denotes that $S$ is a subset of the set complement of $T$.

$\ds S$

$\subseteq$

$\ds \map \complement T$

$\ds \leadstoandfrom \ \ $

$\ds \map \complement {\map \complement T}$

$\subseteq$

$\ds \map \complement S$

$\ds \leadstoandfrom \ \ $

$\ds T$

$\subseteq$

$\ds \map \complement S$

$\blacksquare$

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(f)}$