# Set Difference with Proper Subset

## 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:

$S \setminus T \ne \varnothing$

where $\varnothing$ denotes the empty set.

## Proof

Suppose $S \setminus T = \varnothing$.

Then:

$\not \exists x \in S: x \notin T$
$\forall x \in S: x \in T$

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:

$S \setminus T \ne \varnothing$

$\blacksquare$