Proof for Inclusion-exclusion doesn't make sense

Question

Im currently trying to grasp my professors proof for Inclusion-exclusion, and there is one part of his proof that just doesn't make sense to me and I was hoping for some help here.

Theorem:

Let $A_1,A_2, ... , A_n$ be finite sets.

Then $\displaystyle |\cup_{i=1}^n A_i| = \sum_{i=1}^{k} |A_i|- \sum_{i \neq j}|A_i \cap A_j|+ \sum_{i \neq j \neq k}|A_i \cap A_j \cap A_k|-...+(-1)^{n-1}|A_1 \cap ... \cap A_n|$

Proof:

Lemma needed: Let $n \geq 1$, then $\sum_{k=0}^n (-1)^k {n \choose k}= 0$

Proof Lemma: Apply binomial theorem: $(x+y)^n= \sum_{k=0}^n {n \choose k} x^k y^{n-k}$ with $x=-1, y=1$

Proof of the theorem:

Let $x$ be any element of the union. It suffices to show that $x$ is counted exactly once in the sum.

Suppose $x$ belong to $m$ of the $n$ sets $A_i$, for some $1 \leq m \leq n$. Due to symmetry of formula we can assumed WLOG that $x$ belong to $A_1,...,A_m$.

Part where I don't understand below

Then $x$ is counted $(-1)^{k-1}$ times for each $k$-element subset of ${1,2,...,m}$, $1 \leq k \leq m$.

Hence total number of times $x$ is counted is $$\sum_{k=1}^m (-1)^{k-1} {m \choose k} = - \Big( \sum_{k=0}^m (-1)^k {m \choose k} -(-1)^0 {m \choose k} = -(0-1)=1$$

What does he mean that $x$ is counted $(-1)^{k-1}$ times for each k-element subset? I really don't understand where he get that from.

Answer 1

Let's examine the sums in the inclusion-exclusion principle one at a time.

As in the opener, let $x$ be an arbitrary element that belongs to $A_1, A_2, \dots, A_m$ but not $A_{m+1}, \dots, A_n$.

For the first sum, where we have only $k = 1$ sets involved, we are asking how many times $$ \sum_{i = 1}^n | A_i | $$ will count the particular element $x$. Since $x$ belongs to $m$ of the sets but not the remaining $n - m$, it will be counted precisely $m = \binom{m}{1}$ times.

Stepping up slightly in complexity, for the second sum we are asking how many times $x$ will be counted when we are looking at intersections of $k = 2$ sets. That is, in $$ \sum_{i \neq j} | A_i \cap A_j |, $$ how many times is $x$ counted? Well, if $i$ or $j$ isn't equal to one of $1, 2, \dots, m$, then $x$ won't be counted at all since for $x$ to belong to the intersection $A_i \cap A_j$, it must belong to both $A_i$ and $A_j$. Hence the indices $i \neq j$ that make the count of $x$ go up are precisely the ones where both $i$ and $j$ are from $1, 2, \dots, m$. So the new question becomes: how many ways can we pick two different elements from $1, 2, \dots, m$? This is exactly what binomial coefficients do! They count the ways to pick things from sets, there are $\binom{m}{2}$ ways to pick two different things from $m$ things. Hence this second sum counts $x$ exactly $\binom{m}{2}$ times.

However, the second sum in the inclusion-exclusion principle we are trying to prove has a great big negative sign in front of it, so at this point we have in reality counted $x$ $$ \binom{m}{1} \color{red}{-} \binom{m}{2} $$ times.

Continuing this game, when we are asking about the intersections of $k = 3$ sets, for $x$ to be counted it must belong to each of the three, so this boils down to the amount of ways we can pick three elements from $1, 2, \dots, m$, which is $\binom{m}{3}$.

Keep doing this, and we find in the end that $x$ has been counted $$ \binom{m}{1} \color{red}{-} \binom{m}{2} + \binom{m}{3} \color{red}{-} \binom{m}{4} + \dots $$ times. Notice how (a) the number in the bottom of these binomial coefficients goes up by 1 each time, and (b) the sign alternates from $+$ to $-$ to $+$ and so on. Now if you call the number at the bottom $k$, so that we can attempt to write this sum as something of the shape $$ \sum_{k = 1}^m \color{red}{(\text{something with $k$ that makes the alternating signs})} \binom{m}{k}, $$ what should we put in place of the red stuff? Keep in mind that we want the sign to start with $+$ for $k = 1$.