Intersection with Complement is Empty iff Subset

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Theorem

$S \subseteq T \iff S \cap \map \complement T = \O$

where:

$S \subseteq T$ denotes that $S$ is a subset of $T$
$S \cap T$ denotes the intersection of $S$ and $T$
$\O$ denotes the empty set
$\complement$ denotes set complement.


Corollary

$S \cap T = \O \iff S \subseteq \relcomp {} T$


Proof

\(\ds S\) \(\subseteq\) \(\ds T\)
\(\ds \leadstoandfrom \ \ \) \(\ds S \setminus T\) \(=\) \(\ds \O\) Set Difference with Superset is Empty Set‎
\(\ds \leadstoandfrom \ \ \) \(\ds S \cap \map \complement T\) \(=\) \(\ds \O\) Set Difference as Intersection with Complement

$\blacksquare$


Also presented as

Some sources present this as an alternative definition of a subset:

Set $A$ is a subset of set $B$ if set $A$ contains no member that is not also in set $B$

but this is not mainstream.


Also see


Sources