Inclusion-Exclusion Principle

== Theorem ==
Let $\SS$ be an algebra of sets.
Let $A_1, A_2, \ldots, A_n$ be finite sets.
Let $f: \SS \to \R$ be an additive function.
Then:
=== Corollary ===
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P N$ be the proposition:
$\map P 1$ 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:
Then we need to show:
=== Induction Step ===
This is our induction step:
Consider $\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$.
By the fact that Intersection Distributes over Union, this can be written:
:$\ds \map f {\bigcup_{i \mathop = 1}^r \paren {A_i \cap A_{r + 1} } }$
To this, we can apply the induction hypothesis:
At the same time, we have the expansion of the term $\ds \map f {\bigcup_{i \mathop = 1}^r A_i}$ to take into account.
So we can consider the general term of $s$ intersections in the expansion of $\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$:
:$\ds \paren {-1}^{s - 1} \sum_{\substack {I \subseteq \closedint 1 r \\ \card I = s} } \map f {\bigcap_{i \mathop \in I} A_i} - \paren {-1}^{s - 2} \sum_{\substack {J \subseteq \closedint 1 r \\ \card J = s - 1} } \map f {\bigcap_{i \mathop \in J} A_i \cap A_{r + 1} }$
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:
:$\ds \paren {-1}^{s - 1} \sum_{\substack {I' \subseteq \closedint 1 {r + 1} \\ \card {I'} = s} } \map f {\bigcap_{i \mathop \in I'} A_i}$
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 $\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$.
Just to check, we can see the first term is:
:$\ds \sum_{i \mathop = 1}^r \map f {A_i} + \map f {A_{r + 1} } = \sum_{i \mathop = 1}^{r + 1} \map f {A_i}$
As the expression $\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$ consists only of intersections of two or more elements of $\SS$, we see it does not contribute to this first term.
Finally, let us make sure of the last term.
This is:
:$\ds -\paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }$
which works out as:
:$\ds \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}$
We've done enough.
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
== Examples ==
== Context ==
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 ==
== Sources ==
* : $\S 1.11$: Problems: $12 \ \text{(a)}$
* 
* 
* 
* 
Category:Inclusion-Exclusion Principle
Category:Set Union
Category:Set Systems
Category:Probability Theory
Category:Combinatorics
Category:Named Theorems