Overestimates and underestimates in Inclusion-Exclusion principle

Question

In the celebrated Inclusion-Exclusion principle, $$|\cup A_i|=\sum _i |A_i|-\sum _{i<j} |A_i \cap A_j|+\sum _{i<j<k}|A_i \cap A_j \cap A_k|-\dots +(-1)^{n+1}|\cap A_i|$$

if we take only $m \leq n$ items of the right side, then we would get an overestimate if $m$ is odd, and we obtain an underestimate if $m$ is even, for instance, $$|\cup A_i|\leq \sum _i |A_i|$$ and $$|\cup A_i|\geq \sum _i |A_i|-\sum _{i<j} |A_i \cap A_j|.$$ I want an "elementary" proof for this fact which does not use probability or any measure theory. I tried induction, for the second inequality but could not get anything. I would be thankful for the answer or any leading comment in this connection. Thanks in advance!

Answer 1

Let us introduce some notation. Let $$ S(n,k)=\sum_{1\le i_1<\dots<i_k\le n} |A_{i_1}\cap \dots \cap A_{i_k}| $$ The goal is to prove that $$ \forall m\in \{1,\dots,n\},\quad \begin{matrix} \text{$m$ is odd}\implies\left|\displaystyle\bigcup_{i=1}^n A_i\right|\le S(n,1)-S(n,2)\dots +S(n,m)\\ \text{$m$ is even}\implies\left|\displaystyle\bigcup_{i=1}^n A_i\right|\ge S(n,1)-S(n,2)\dots -S(n,m) \end{matrix}\tag{$*$} $$ The proof is by induction on $n$. The base case $n=1$ is obvious. So, let us assume that $(*)$ is true for $n-1$, and we will prove it for $n$.

Let $m\in \{1,\dots,n\}$ be given. Assume for now that $m$ is odd (the even case is analogous). We start with $$ \left|\bigcup_{i=1}^n A_i\right|=|A_n|+ \left|\bigcup_{i=1}^{n-1} A_i\right|-\left|\bigcup_{i=1}^{n-1} (A_i \cap A_n)\right|\tag1 $$ This is just $|E\cup F|=|E|+|F|-|E\cap F|$ applied with $E=A_n$ and $F=\bigcup_{i=1}^{n-1}A_n$.

Now, we apply the induction hypothesis twice. First, we apply it to $\left|\bigcup_{i=1}^{n-1} A_i\right|$, using the same $m$, getting $$ \left|\bigcup_{i=1}^{n-1} A_i\right|\le S(n-1,1)-S(n-1,2)\dots +S(n-1,m)\tag2 $$ Next, we apply the induction hypothesis to the other union $\left|\bigcup_{i=1}^{n-1} (A_i \cap A_n)\right|$. However, this union is being subtracted, so we instead need a lower bound, so we use $m-1$ instead of $m$. That is, $$ \left|\bigcup_{i=1}^{n-1} (A_i \cap A_n)\right|\ge T(n-1,1)-T(n-1,2)+\dots -T(n-1,m-1)\tag3 $$ where we define $$ \begin{align} T(n-1,k) &:=\sum_{1\le i_1< \dots <i_k\le n-1} |(A_{i_1}\cap A_n)\cap (A_{i_2}\cap A_n)\cap \dots \cap (A_{i_k}\cap A_n)| \\ T(n-1,0) &:= 0 \end{align} $$ Combining $(1)$, $(2)$ and $(3)$, we get $$ \left|\bigcup_{i=1}^{n} A_i\right|\le \sum_{k=1}^{m} (-1)^{k-1}\Big(S(n-1,k)+T(n-1,k-1)\Big)\tag4 $$ To complete the proof, you just need to realize that $$ \forall k\in \{1,\dots,n\}\qquad S(n,k)=S(n-1,k)+T(n-1,k-1)\tag5 $$ This is true because $S(n-1,k)$ consists of the summands in $S(n,k)$ which do not involve $A_n$, while $T(n-1,k-1)$ is the summands in $S(n,k)$ which do involve $A_n$. The only exception to this is the case $k=1$; here, we instead use $S(n,1)=S(n-1,1)+|A_n|$, which is valid because there is an $|A_n|$ floating out front in $(1)$.

Anyways, if you apply $(5)$ to $(4)$, then $(4)$ becomes $(*)$ exactly. QED (at least when $m$ is odd; when $m$ is even, just flip all inequality signs).

Answer 2

It is enough to prove the inequalities of indicators $$1_{A} \leq \sum_{i}1_{A_i},$$ $$1_{A} \geq \sum_{i}1_{A_i} - \sum_{i < j}1_{A_i \cap A_j},$$ $$1_{A} \leq \sum_{i}1_{A_i} - \sum_{i < j}1_{A_i \cap A_j} + \sum_{i < j < k}1_{A_i \cap A_j \cap A_k},$$ etc. Because if these inequalities are shown, then evaluating and summing them over all $\omega \in A_1 \cup \dots \cup A_n$ finishes the proof.

So suppose $\omega \in A_1 \cup \dots \cup A_n$ is arbitray. So $\omega$ is in $k \geq 1$ of the sets. Then for odd $m$, we want to show that $$1 \leq k - \binom{k}{2} + \binom{k}{3} - \dots \pm \binom{k}{m},$$ and for even $m$ we want to show the reverse inequality. So we want to show that $$\sum_{i = 0}^{m}(-1)^i\binom{k}{i} \leq 0$$ for odd $m$, and the reverse inequality for even $m$. But this follows from the formula $$\sum_{i = 0}^{m}(-1)^i\binom{k}{i} = (-1)^m\binom{k - 1}{m}.$$

Answer 3

Let $A = A_1 \cup \cdots \cup A_n$. We first partition $A$ according to the value of $(\mathbf{1}_{A_i}, \ldots, \mathbf{1}_{A_n})$. More specifically, for each $\varepsilon \in \{ 0, 1\}^n$ we define $A_{\varepsilon}$ as

\begin{align*} A_{\varepsilon} = \{ \omega \in A : \mathbf{1}_{A_i}(\omega) = \varepsilon_i \text{ for all } i = 1, \ldots, n\} = \bigcap_{i=1}^{n} A_{i,\varepsilon_i}, \end{align*}

where $A_{i,1} = A_i$ and $A_{i,0} = A\setminus A_i$ for each $i = 1, \ldots, n$. In other words, $A_{\varepsilon}$ is the set of all $\omega \in A$ such that $[\omega \in A_i \iff \varepsilon_i = 1]$ holds for all $i = 1, \ldots, n$. Then it is clear that

$$ A_j = \{ \omega \in A : \mathbf{1}_{A_j}(\omega) = \varepsilon_j\} = \bigcup_{\substack{\varepsilon \in \{0, 1\}^n \\ \varepsilon_j = 1}} A_{\varepsilon}. $$

Using this, we can rewrite the (unsigned) $k$-th term $S_k$ in the inclusion-exclusion formula as:

\begin{align*} S_k := \sum_{i_1 < \ldots < i_k} |A_{i_1} \cap \cdots \cap A_{i_k}| &= \sum_{i_1 < \ldots < i_k} \Biggl( \sum_{\varepsilon \in \{0, 1\}^n} |A_{\varepsilon}| \cdot \mathbf{1}_{\{\varepsilon_{i_1} = 1, \ldots, \varepsilon_{i_k} = 1\}} \Biggr) \\ &= \sum_{\varepsilon \in \{0, 1\}^n} |A_{\varepsilon}| \Biggl( \sum_{i_1 < \ldots < i_k} \mathbf{1}_{\{\varepsilon_{i_1} = 1, \ldots, \varepsilon_{i_k} = 1\}} \Biggr) . \end{align*}

The parenthesized sum in the last line counts the number $k$-subsets of $\{i : \varepsilon_i = 1\}$, and so, it is evaluated as $\binom{|\varepsilon|}{k}$, where $|\varepsilon| = \varepsilon_1 + \cdots + \varepsilon_n$. So, the above double sum for $S_k$ is reduced to

\begin{align*} S_k &= \sum_{\varepsilon \in \{0, 1\}^n} \binom{|\varepsilon|}{k} |A_{\varepsilon}|. \tag{1} \end{align*}

Using this, the sum of first $m$ terms in the inclusion-exclusion principle is given by

$$ \sum_{k=1}^{m} (-1)^{k-1} S_k = \sum_{\varepsilon \in \{0, 1\}^n} \Biggl[ \sum_{k=1}^{m} (-1)^k \binom{|\varepsilon|}{k} \Biggr] |A_{\varepsilon}|. \tag{2} $$

To make use of this equality, we note that

\begin{align*} \sum_{k=1}^{m} (-1)^{k-1} \binom{r}{k} &= \sum_{k=1}^{m} (-1)^{k-1} \left[ \binom{r-1}{k-1} + \binom{r-1}{k} \right] \\ &= 1 + (-1)^{m-1}\binom{r-1}{m} \ \begin{cases} \geq 1, & \text{if $m$ is odd,} \\ \leq 1, & \text{if $m$ is even.} \end{cases} \tag{3} \end{align*}

So by plugging $\text{(3)}$ to $\text{(2)}$, we conclude:

\begin{align*} \text{$m$ odd} &\quad\implies\quad \sum_{k=1}^{m} (-1)^{k-1} S_k \geq \sum_{\varepsilon \in \{0, 1\}^n} |A_{\varepsilon}| = |A|, \\ \text{$m$ even} &\quad\implies\quad \sum_{k=1}^{m} (-1)^{k-1} S_k \leq \sum_{\varepsilon \in \{0, 1\}^n} |A_{\varepsilon}| = |A|. \end{align*}