Does inclusion-exclusion formula holds for coutanble index set? Does inclusion-exclusion formula holds for countable index set?
Here is the formula for index set of size 2.
\begin{align}
P(A \cup B)=P(A)+P(B)-P(A\cap B)
\end{align}
 A: There are formulas for a finite index set, which aren't the same as the one you have there, iirc for three sets it is
$P(A \cup B \cup C)=P(A) + P(B)+P(C)+P(A\cap B\cap C) - P(A \cap B) - P(B\cap C) - P(C \cap A)$
For an infinite index set though this process may or may not converge. Take for example the set of integers. For reference, the formula for arbitrary n is given as;
$P(\cup_{i=1}^nA_n) = \sum_{i=1}^n(-1)^{i-1}\mathbb{P(i)}$
Where $\mathbb{P}(i)$ is the sum of P of every union of i elements.
A: According to definition of the probability space $(\Omega,\mathcal{F},P)$ you can see that 


*

*$\mathcal{F}$ is closed under countable unions.

*$P$ is defined for all elements of $\mathcal{F}$.

*$P$ is additive for union of disjoint elements of $\mathcal{F}$.
The colloraly of the definition and the De Morgan Laws is:


*$\mathcal{F}$ is closed under countable intersections.


Let us consider a series $A_i$ where $A_i\in\mathcal{F}$, $i=1,2\dots$. The probability of the union can be constructed recursively as follows:
First step is clear
$$
P(\cup_{i=1}^1 A_i) = P(A_1)
$$
and we know for it that $A_1\in\mathcal{F}$.
The recursive step is
$$
P(\cup_{i=1}^n A_i) = P(\cup_{i=1}^{n-1} A_i) + P(A_n)-P([\cup_{i=1}^{n-1} A_i]\cap A_n)
$$
The challenge is whether this $(\cup_{i=1}^n A_i)\in\mathcal{F}$. We can resolve it as follows: from the recursion we know that $(\cup_{i=1}^{n-1} A_i)\in\mathcal{F}$ and from problem formulation that $A_n\in\mathcal{F}$. Since $\mathcal{F}$ is closed under intersections, we know that $([\cup_{i=1}^{n-1} A_i]\cap A_n)\in\mathcal{F}$. Thus, the recursive step is well defined. Repeating it till infinity, you obtain the calculation for any countable set.
