Dirichlet's Box Principle

Theorem
Let $S$ be a finite set whose cardinality is $n$.

Let $S_1, S_2, \ldots, S_k$ be a partition of $S$ into $k$ subsets.

Then:
 * at least one subset $S_i$ of $S$ contains at least $\ceiling {\dfrac n k}$ elements

where $\ceiling {\, \cdot \,}$ denotes the ceiling function.

Proof
no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements.

Then the maximum number of elements of any $S_i$ would be $\ceiling {\dfrac n k} - 1$.

So the total number of elements of $S$ would be no more than $k \paren {\ceiling {\dfrac n k} - 1} = k \ceiling {\dfrac n k} - k$.

There are two cases:
 * $n$ is divisible by $k$
 * $n$ is not divisible by $k$.

Suppose $k \divides n$.

Then $\ceiling {\dfrac n k} = \dfrac n k$ is an integer and:
 * $k \ceiling {\dfrac n k} - k = n - k$

Thus:
 * $\displaystyle \card S = \sum_{i \mathop = 1}^k \card {S_i} \le n - k < n$

This contradicts the fact that $\card S = n$.

Hence our assumption that no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements was false.

Next, suppose that $k \nmid n$.

Then:
 * $\card S = k \ceiling {\dfrac n k} - k < \dfrac {k \paren {n + k} } k - k = n$

and again this contradicts the fact that $\card S = n$.

In the same way, our assumption that no subset $S_i$ of $S$ has as many as $\ceiling {\dfrac n k}$ elements was false.

Hence, by Proof by Contradiction, there has to be at least $\ceiling {\dfrac n k}$ elements in at least one $S_i \subseteq S$.

Source of Name
It is known as the Pigeonhole Principle because of the following.

Suppose you have $n + 1$ pigeons, but have only $n$ holes for them to stay in.

By the Pigeonhole Principle at least one of the holes houses $2$ pigeons.

Also see

 * Infinite Ramsey's Theorem