Inclusion-Exclusion Principle

Theorem
Let $\mathcal S$ be an algebra of sets.

Let $A_1, A_2, \ldots, A_n$ be finite sets.

Let $f: \mathcal S \to \R$ be an additive function.

Then:

Corollary
Let $\mathcal S$ be an algebra of sets.

Let $A_1, A_2, \ldots, A_n$ be finite sets which are pairwise disjoint.

Let $f: \mathcal S \to \R$ be an additive function.

Then:


 * $\displaystyle f \left({\bigcup_{i=1}^n A_i}\right) = \sum_{i=1}^n f \left({A_i}\right)$

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$P(1)$ is true, as this just says $f \left({A_1}\right) = f \left({A_1}\right)$.

Basis for the Induction
$P(2)$ is the case:
 * $f \left({A_1 \cup A_2}\right) = f \left({A_1}\right) + f \left({A_2}\right) - f \left({A_1 \cap A_2}\right)$

which is the result Additive Function on Union of Sets.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r+1}\right)$ is true.

So this is our induction hypothesis:

Then we need to show:

Induction Step
This is our induction step:

Consider $\displaystyle f \left({\bigcup_{i=1}^r A_i \cap A_{r+1}}\right)$.

By the fact that Intersection Distributes over Union, this can be written:


 * $\displaystyle f \left({\bigcup_{i=1}^r \left({A_i \cap A_{r+1}}\right)}\right)$

To this, we can apply the induction hypothesis:

At the same time, we have the expansion of the term $\displaystyle f \left({\bigcup_{i=1}^r A_i}\right)$ to take into account.

So we can consider the general term of $s$ intersections in the expansion of $\displaystyle f \left({\bigcup_{i=1}^{r+1} A_i}\right)$:


 * $\displaystyle \left({-1}\right)^{s-1} \sum_{{i \in I} \atop {\left|{I}\right| = s}} f \left({\bigcap_{i \in I} A_i}\right) - \left({-1}\right)^{s-2} \sum_{{i \in J} \atop {\left|{J}\right| = s-1}} f \left({\bigcap_{i \in J} A_i \cap A_{r+1}}\right)$

where:
 * $I$ ranges over all sets of $s$ elements out of $\left[{1 .. r}\right]$
 * $J$ ranges over all sets of $s-1$ elements out of $\left[{1 .. r}\right]$
 * $1 \le s \le r$

Messy though it is, it can be seen that this expression is merely:


 * $\displaystyle \left({-1}\right)^{s-1} \sum_{{i \in I'} \atop {\left|{I'}\right| = s}} f \left({\bigcap_{i \in I'} A_i}\right)$

where this time, $I'$ ranges over all sets of $s$ elements out of $\left[{1 \,. \, . \, r+1}\right]$.

This is the required term in $s$ intersections in the expansion of $\displaystyle f \left({\bigcup_{i=1}^{r+1} A_i}\right)$.

Just to check, we can see the first term is:
 * $\displaystyle \sum_{i=1}^r f \left({A_i}\right) + f \left({A_{r+1}}\right) = \sum_{i=1}^{r+1} f \left({A_i}\right)$

As the expression $\displaystyle f \left({\bigcup_{i=1}^r A_i \cap A_{r+1}}\right)$ consists only of intersections of two or more elements of $\mathcal S$, we see it does not contribute to this first term.

Finally, let us make sure of the last term - this is:
 * $\displaystyle - \left({-1}\right)^{r-1} f \left({\bigcap_{i=1}^r A_i \cap A_{r+1}}\right)$

which works out as:
 * $\displaystyle \left({-1}\right)^r f \left({\bigcap_{i=1}^{r+1} A_i}\right)$

We've done enough.

So $P \left({r}\right) \implies P \left({r+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

Proof of Corollary
As $A_1, A_2, \ldots, A_n$ are pairwise disjoint, their intersections are all empty.

The main result holds, but from Cardinality of Empty Set, all the terms apart from the first vanish.

Comment
This result is usually quoted in the context of combinatorics, where $f$ is the cardinality function.

It is also seen in the context of probability theory, in which $f$ is taken to be a probability measure.

Historical Notes
This formula, in various forms, has been attributed to:
 * Abraham de Moivre
 * Daniel da Silva
 * Joseph Sylvester
 * Henri Poincaré.