Set Complement inverts Subsets/Proof 2

Theorem

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

$\ds S$

$\subseteq$

$\ds T$

$\ds \leadstoandfrom \ \ $

$\ds (x \in S$

$\implies$

$\ds x \in T)$

Definition of Subset

$\ds \leadstoandfrom \ \ $

$\ds (x \notin T$

$\implies$

$\ds x \notin S)$

$\ds \leadstoandfrom \ \ $

$\ds (x \in \map \complement T$

$\implies$

$\ds x \in \map \complement S)$

Definition of Set Complement

$\ds \map \complement T$

$\subseteq$

$\ds \map \complement S$

Definition of Subset

$\blacksquare$

1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 6.7$: Subsets