Set Difference with Proper Subset
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:
- $S \setminus T \ne \O$
where $\O$ denotes the empty set.
Proof
Aiming for a contradiction, suppose $S \setminus T = \O$.
Then:
- $\not \exists x \in S: x \notin T$
By De Morgan's laws:
- $\forall x \in S: x \in T$
By definition of subset:
- $S \subseteq T$
By definition of proper subset, we have that $T \subseteq S$ such that $T \ne S$.
But we have $T \subseteq S$ and $S \subseteq T$.
So by definition of set equality:
- $S = T$
From this contradiction it follows that:
- $S \setminus T \ne \O$
$\blacksquare$