# Inclusion / exclusion principle theorem form

I'm having trouble understanding how this form of the principle (on wiki) results in the form below.

Wiki form:

Using three sets:

My confusion is the last $$(-1)^{n-1} |A_1 \cap \cdots \cap A_n|$$. Where does this last form appear in the three set example?

Also, if using two sets vs three, is the $$\sum_{1 \le i \lt j \lt k \le n}$$ unsatisfiable because there are no three $$i, j, k$$ terms to satisfy it? In that case this summation term disappears?

• Using three sets, you have $n=3$, so the term $+|A\cap B\cap C|$ is precisely the term $(-1)^{3-1}|A_1\cap A_2 \cap A_3|$ where $A_1=A,A_2=B,A_3=C$ are the assigned names, as you can see from the LHS. For your second question, of course the $\sum_{1\le i<j<k\le n} |A_i\cap A_j \cap A_k|$ term is not there, because this term asks you to take intersection of three distinct sets, which you don't even have. For the two set example, you have the first term and the second term (the second term is obviously the last term as well, since $n=2$). – Fawkes4494d3 Sep 1 at 16:53
• Your question/complaint effectively the same as complaining about the $n$'th triangular number being written as $T(n)=1+2+3+\dots + n$. It is well accepted that this is $T(1)=1, ~T(2)=1+2, ~T(3)=1+2+3,~T(4)=1+2+3+4,\dots$ and that when $n$ happens to be small enough that you simply don't include the irrelevant terms, those terms who are larger than $n$. Yes, it could have instead been written $T(n)=\sum\limits_{k=1}^n k$ and had no ambiguity for if $n$ were $1,2,3$ but the usage of ellipses like here is common enough that the intended meaning is generally accepted. – JMoravitz Sep 1 at 17:01
• I suggest reading How acceptable is an ellipsis in formal mathematics? – JMoravitz Sep 1 at 17:04
• My confusion was a simple one: the last summation term in which the ellipsis appeared confused me. Now I understand that it's shown with ellipsis and that the sign alternates. Thanks! – Nick Sep 1 at 17:15

Rename $$A,B$$, and $$C$$ as $$A_1,A_2$$, and $$A_3$$, respectively. Then

$$\sum_{i=1}^n|A_i|=|A_1|+|A_2|+|A_3|\,,$$

$$\sum_{i\le i

and

\begin{align*} (-1)^{n-1}|A_1\cap\ldots\cap A_n|&=(-1)^{3-1}|A_1\cap A_2\cap A_3|\\ &=|A_1\cap A_2\cap A_3|\,. \end{align*}

The general form is written for more than $$3$$ sets; it should be understood as

$$\left|\bigcup_{i=1}^nA_i\right|=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{i+1}\left|\bigcap_{i\in I}A_i\right|\,,$$

where $$[n]=\{1,2,\ldots,n\}$$.

$$n=3$$ and the last term in the expansion is $$(-1)^{3-1}|A \cap B \cap C| = |A \cap B \cap C|$$

It's confusing since the form is actually written out for $$n>3$$, but for $$n=3$$, the last term and the next-to-last are the same...