Set Difference with Proper Subset is Proper Subset

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a set.

Let $T \subsetneq S$ be a proper subset of $S$.

Let $S \setminus T$ denote the set difference between $S$ and $T$.


Then:

$S \setminus T$ is a proper subset of $S$


Proof

From Set Difference is Subset:

$S \setminus T \subseteq S$

From Set Difference with Proper Subset:

$S \setminus T \ne \O$

By definition of proper subset:

$T \ne \O$

From Intersection with Subset is Subset:

$S \cap T = T$

Hence:

$S \cap T \ne \O$

From the contrapositive statement of Set Difference with Disjoint Set:

$S \setminus T \ne S$


It follows that $S \setminus T$ is a proper subset by definition.

$\blacksquare$