Connection between probability and finite cardinality for Bonferroni inequalities

Question

Let

$$S_1=\sum_i|A_i|, \quad S_2= \sum_{i<j}|A_i \cap A_j|, \quad S_3=\sum_{i<j<k} |A_i \cap A_j \cap A_k| ,\quad \cdots$$

Define alternating partial sums $$R_1 = S_1,\quad R_2 = S_1 - S_2, \quad R_3 = S_1 - S_2 + S_3,\quad \cdots $$

I've long believed but never seen explicitly union bound / Bonferroni-type inequalities for finite cardinality sets:

$$R_2 \le R_4 \le \cdots \le \left|\bigcup_i A_i \right| \le \cdots \le R_3 \le R_1 $$

Can these be proved from the probability formulation? $|A_i \cup A_j| \le |A_i| + |A_j|$ just as $P(A_i \cup A_j) \le P(A_i) + P(A_j)$. My idea was define a discrete uniform distribution $U[1,n]$. Let $B_i$ be subsets of the domain and $A_i = x \in B_i$ for $x \sim U[1,n]$. Then $P(A_i) = |B_i|/n$ (I'm not sure if this notation is correct). Scaling by $n$, this should give the above bounds? Actually, can this be proved from first principles just using induction on inclusion-exclusion?

Answer 1

I know how to prove $R_{2n}\le |\bigcup A_i|\le R_{2n-1}$ for each $n\ge 1$, but I do not know how to prove $R_2\le R_4\le \dots \le R_3\le R_1$.

Let $\Omega$ be a set containing $\bigcup_{i=1}^n A_i$. If you choose an element $\omega$ from $\Omega$ uniformly at random, the probability that $\omega\in A_i$ is $|A_i|/|\Omega|$. Now, let $E_i$ be the event that $\omega\in A_i$. If you take the inequalities you want to prove and multiply through by $1/|\Omega|$, then the result is exactly the same thing, where all sets $A_i$ are replaced with events $E_i$ and all cardinalities $|\cdot|$ are replaced with probabilities. Therefore, the probabilistic Bonferroni inequalities imply the set-based ones, and vice versa.

There are some slick proofs of the Bonferroni inequalities at How to prove Bonferroni inequalities? and Overestimates and underestimates in Inclusion-Exclusion principle, including some inductive proofs from first principles.