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I have difficult reading this notation

$\left| \bigcup_{i=1}^{n} A_{i} \right|=\sum_{k=1}^{n}(-1)^{k-1}\sum_{I\in\binom{{1,2,3,...n}}{k}} \left|\bigcap_{i=I}A_{i} \right|$

could anyone work out an expansion up to n=3 for me?

2 Answers 2

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It is a little bit odd the way you wrote it, i would write it like $$\left| \bigcup_{i=1}^{n} A_{i} \right|=\sum_{k=1}^{n}(-1)^{k-1}\sum_{I\in\binom{\color{red}{\{1,2,3,...n\}}}{k}} \left|\bigcap_{i\color{red}{\in} I}A_{i} \right|,$$ being that said, remember that $$I\in \binom{\{1,\cdots , n\}}{k},$$ means that $I\subseteq [n]$ and $|I|=k.$ So take $n=3,k=2$ then the possible sets are $I=\{1,2\},\{1,3\},\{2,3\}$ so $$|\bigcap _{i\in I} A_i|=|A_1\cap A_2|,|A_1\cap A_3|,|A_2\cap A_3|$$ respectively. I will let you write it for $n=3$. For $n=2$ then is as follows $$\left| \bigcup_{i=1}^{2} A_{i} \right|=\sum_{k=1}^{2}(-1)^{k-1}\sum_{I\in\binom{\color{red}{\{1,2\}}}{k}} \left|\bigcap_{i\color{red}{\in} I}A_{i} \right|=(-1)^{1-1}(\underbrace{|A_1|}_{\text{relative to $\{1\}$}}+\underbrace{|A_2|}_{\text{relative to $\{2\}$}})+(-1)^{2-1}\underbrace{|A_1 \cap A_2|}_{\text{relative to $\{1,2\}$}}.$$

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    @ Phicar Thanks! pretty good help. For n=3 – 2017-01-27
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    it is too long to type in comment, so I wrote it as an answer. please take a look and see if it is correct.! – 2017-01-27
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    @Tmm Yes, it is correct. You forgot a parenthesis tho and the left part you need to replace $\{1,2\}$ by $\{1,2,3\}$ and the upper limit of the sum. – 2017-01-27
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    @ Phicar Then how about$ \left| \bigcup_{i=1}^{n} A_{i}\right|= \sum_{\emptyset ≠ I \subseteq \{1,2,....n\}} (-1)^{\lvert I \rvert -1} \left| \bigcap_{i \in I}A_{i}\right|$ It does not seem to have anything indicating the n chooses k? – 2017-01-27
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    @Tmm it is just the full expression . The one you wrote in your question is grouping(by associativity and factoring out the $(-1)^k$) the terms of the same cardinality. (subsets of same cardinality is when $\binom{\{1,\cdots , n\}}{k}$ appears) – 2017-01-27
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    Ok, I missed the knowledge of cardinality, thanks. it all makes sense now. It's just shuffled everything as subsets of the set {1,2,....n} without ordering it. But it makes the formula looks more handsome and simple. right? – 2017-01-27
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    Yes, it looks simpler. – 2017-01-27
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$$\left| \bigcup_{i=1}^{3} A_{i} \right|=\sum_{k=1}^{2}(-1)^{k-1}\sum_{I\in\binom{\color{red}{\{1,2\}}}{k}} \left|\bigcap_{i\color{red}{\in} I}A_{i} \right| =(-1)^{1-1} (\underbrace{|A_1|}_{\text{relative to $\{1\}$}} +\underbrace{|A_2|}_{\text{relative to $\{2\}$}} + \underbrace{|A_{3}|}_{\text{relative to {3}}}) +(-1)^{2-1} (\underbrace{|A_1 \cap A_2|}_{\text{relative to $\{1,2\}$}}+ \underbrace{|A_1 \cap A_2|}_{\text{relative to $\{2,3\}$}}+ \underbrace{|A_1 \cap A_2|}_{\text{relative to $\{1,3\}$}})+ (-1)^{3-1}(\underbrace{|A_1 \cap A_2 \cap A_3|}_{\text{relative to $\{1,2,3\}$}}).$$