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 $$\left \lceil {n/k} \right \rceil$$ elements.

It can also be seen in the simpler form:

If a set of $$n$$ distinct objects is partitioned into $$k$$ subsets, where $$0 < k < n$$, then at least one subset must contain at least two elements.

Proof
Suppose this were not the case, and no subset $$S_i$$ of $$S$$ has as many as $$\left \lceil {n/k} \right \rceil$$ elements.

Then the maximum number of elements of any $$S_i$$ would be $$\left \lceil {n/k} \right \rceil - 1$$.

So the total number of elements of $$S$$ would be no more than $$k \left({\left \lceil {n/k} \right \rceil - 1}\right) = k \left \lceil {n/k} \right \rceil - k$$.

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

Suppose $$k \backslash n$$.

Then $$\left \lceil {\frac n k} \right \rceil = \frac n k$$ is an integer and $$k \left \lceil {n/k} \right \rceil - k = n - k$$.

Thus $$\sum_{i=1}^k \left \vert {S_i}\right \vert \le n-k < n$$.

This contradicts our assumption that no subset $$S_i$$ of $$S$$ has as many as $$\left \lceil {n/k} \right \rceil$$ elements.

Next, suppose that $$k \nmid n$$. Then $$k \left \lceil {n/k} \right \rceil - k < \frac {k \left({n+1}\right) - k}{k} = n$$ and again this contradicts our assumption that no subset $$S_i$$ of $$S$$ has as many as $$\left \lceil {n/k} \right \rceil$$ elements.

Either way, there has to be at least $$\left \lceil {n/k} \right \rceil$$ elements in at least one $$S_i \subseteq S$$.

Historical Note
This principle has been attributed to Dirichlet.