# 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:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i}$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \map f {A_i}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}$ $\displaystyle$  $\, \displaystyle \cdots \,$ $\displaystyle$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}$

### 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:

$\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$

## Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P N$ be the proposition:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i}$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \map f {A_i}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}$ $\displaystyle$  $\, \displaystyle \cdots \,$ $\displaystyle$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}$

$\map P 1$ is true, as this just says $\map f {A_1} = \map f {A_1}$.

### Basis for the Induction

$\map P 2$ is the case:

$\map f {A_1 \cup A_2} = \map f {A_1} + \map f {A_2} - \map f {A_1 \cap A_2}$

which is the result Additive Function is Strongly Additive.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.

So this is our induction hypothesis:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^r A_i}$ $=$ $\displaystyle \sum_{i \mathop = 1}^r \map f {A_i}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le r} \map f {A_i \cap A_j}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r} \map f {A_i \cap A_j \cap A_k}$ $\displaystyle$  $\, \displaystyle \cdots \,$ $\displaystyle$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i}$

Then we need to show:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$ $=$ $\displaystyle \sum_{i \mathop = 1}^{r + 1} \map f {A_i}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le r + 1} \map f {A_i \cap A_j}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r + 1} \map f {A_i \cap A_j \cap A_k}$ $\displaystyle$  $\, \displaystyle \cdots \,$ $\displaystyle$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}$

### Induction Step

This is our induction step:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$ $=$ $\displaystyle \map f {\bigcup_{i \mathop = 1}^r A_i \cup A_{r + 1} }$ $\displaystyle$ $=$ $\displaystyle \map f {\bigcup_{i \mathop = 1}^r A_i} + \map f {A_{r + 1} } - \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$ Basis for the Induction

Consider $\displaystyle \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$.

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

$\displaystyle \map f {\bigcup_{i \mathop = 1}^r {A_i \cap A_{r + 1} } }$

To this, we can apply the induction hypothesis:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^r {A_i \cap A_{r + 1} } }$ $=$ $\displaystyle \sum_{i \mathop = 1}^r \map f {A_i \cap A_{r + 1} }$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le r} \map f {A_i \cap A_j \cap A_{r + 1} }$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r} \map f {A_i \cap A_j \cap A_k \cap A_{r + 1} }$ $\displaystyle$  $\, \displaystyle \cdots \,$ $\displaystyle$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }$

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

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

$\displaystyle \paren {-1}^{s - 1} \sum_{\substack {I \subseteq \closedint 1 r \\ \card I = s} } \map f {\bigcap_{i \mathop \in I} A_i} - \paren {-1}^{s - 2} \sum_{\substack {J \subseteq \closedint 1 r \\ \card J = s - 1} } \map f {\bigcap_{i \mathop \in J} A_i \cap A_{r + 1} }$

where:

$I$ ranges over all sets of $s$ elements out of $\closedint 1 r$
$J$ ranges over all sets of $s - 1$ elements out of $\closedint 1 r$
$1 \le s \le r$

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

$\displaystyle \paren {-1}^{s - 1} \sum_{\substack {I' \subseteq \closedint 1 {r + 1} \\ \card {I'} = s} } \map f {\bigcap_{i \mathop \in I'} A_i}$

where this time, $I'$ ranges over all sets of $s$ elements out of $\closedint 1 {r + 1}$ and $1 \le s \le r + 1$.

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

Just to check, we can see the first term is:

$\displaystyle \sum_{i \mathop = 1}^r \map f {A_i} + \map f {A_{r + 1} } = \sum_{i \mathop = 1}^{r + 1} \map f {A_i}$

As the expression $\displaystyle \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$ 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:

$\displaystyle -\paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }$

which works out as:

$\displaystyle \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}$

We've done enough.

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

Therefore:

 $\displaystyle \map f {\bigcup_{i \mathop = 1}^n A_i}$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \map f {A_i}$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}$ $\displaystyle$  $\, \displaystyle \cdots \,$ $\displaystyle$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}$

$\blacksquare$

## 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 Note

The Inclusion-Exclusion Principle, in various forms, has been attributed to: