# Cardinality of Set Difference

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## Theorem

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

Then:

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

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

## Proof

From Intersection is Subset:

- $S \cap T \subseteq S$
- $S \cap T \subseteq T$

From Subset of Finite Set is Finite:

- $S \cap T$ is finite.

We have:

\(\displaystyle \card {S \setminus T}\) | \(=\) | \(\displaystyle \card {S \setminus \paren {S \cap T} }\) | Set Difference with Intersection is Difference | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \card S - \card {S \cap T}\) | Cardinality of Set Difference with Subset |

$\blacksquare$