# Cardinality of Set Difference with Subset

## Theorem

Let $S$ and $T$ be sets such that $T$ is finite.

Let $T \subseteq S$.

Then:

$\card {S \setminus T} = \card S - \card T$

where $\card S$ denotes the cardinality of $S$.

## Proof

$T \subseteq S \iff T \setminus S = \O$
$S = \paren {S \setminus T} \cup T$

Thus from Cardinality of Set Union:

$\card S = \card T + \card {S \setminus T} - \card {T \cap \paren {S \setminus T} }$
$T \cap \paren {S \setminus T} = 0$

Hence the result.

$\blacksquare$