Fundamental Principle of Counting

Theorem
Let $A$ be a set.

Let $\left \langle {B_n} \right \rangle$ be a sequence of distinct finite subsets of $A$ which form a partition of $A$.

Let $p_k = \left|{B_k}\right|$ for each $k \in \left[{1 \,.\,.\, n}\right]$.

Then:
 * $\displaystyle \left|{A}\right| = \sum_{k=1}^n p_k$

That is, the sum of the numbers of elements in the subsets of a partition of a set is equal to the total number of elements in the set.

Proof
Let $r_0 = 0$, and let:


 * $\displaystyle \forall k \in \left[{1 \,.\,.\, n}\right]: r_k = \sum_{j=1}^k {p_j}$

Then $r_{k-1} + p_k = r_k$, so $r_{k-1} < r_k$.

Thus by Isomorphism to Closed Interval, $\left[{r_{k-1} \,.\,.\, r_k}\right]$ has $r_k - r_{k-1} = p_k$ elements.

As a consequence, there exists a bijection $\sigma_k: B_k \to \left[{r_{k-1} \,.\,.\, r_k}\right]$ for each $k \in \left[{1 \,.\,.\, n}\right]$.

Let $\sigma: A \to \N$ be the mapping that satisfies:


 * $\forall x \in B_k: \forall k \in \N: \sigma \left({x}\right) = \sigma_k \left({x}\right)$

By Strictly Increasing Sequence on Ordered Set, $\left \langle {r_k} \right \rangle_{0 \le k \le n}$ is a strictly increasing sequence of natural numbers.

Thus by Strictly Increasing Sequence induces Partition, $\sigma: A \to \left[{1 \,.\,.\, r_n}\right]$ is a bijection.

By Isomorphism to Closed Interval, $\left[{1 \,.\,.\, r_n}\right]$ has $r_n$ elements.

Hence the result.