Intersection with Complement is Empty iff Subset

From ProofWiki
Jump to navigation Jump to search


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


$S \subseteq T$ 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.


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


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


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