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

From Set Difference with Superset is Empty Set:

$T \subseteq S \iff T \setminus S = \O$

From Set Difference and Intersection form Partition:

$S = \paren {S \setminus T} \cup T$

Thus from Cardinality of Set Union:

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

But from Set Difference Intersection with Second Set is Empty Set:

$T \cap \paren {S \setminus T} = 0$

Hence the result.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 1.6$. Difference and complement: Example $20$