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I'd like to express every combination represented using union or intersection between multiple sets.

For example, we have 3 sets $A_1,A_2,A_3$, the following combinations can be generated by unions and/or intersections between $A_1, A_2, A_3$.

$$A_1 \setminus (A_2 \cup A_3)$$ $$A_2 \setminus (A_1 \cup A_3)$$ $$A_3 \setminus (A_1 \cup A_2)$$ $$(A_1 \cap A_2) \setminus A_3$$ $$(A_1 \cap A_3) \setminus A_2$$ $$(A_2 \cap A_3) \setminus A_1$$ $$A_1 \cap A_2 \cap A_3$$

Using Venn diagram, the combination means the distinct area enclosed by lines of $A_1, A_2, A_3$.

I'd like to formalize the combination for multiple sets $A_i \; (1 \leq i \leq n)$. Then I considered the following expression, but I cannot prove the expression is true.

$$\bigcap_{r \in N} A_r \setminus \bigcup_{s \in N^\complement} A_s \quad (\emptyset\not=N\subseteq\{1,2,\dots,n\}, N^\complement=\{1,2,\dots,n\}\setminus N)$$

If the expression is true, please prove it. If not, please correct the expression to represent all combination of multiple sets.

In addition, I think that the cardinality of union of multiple sets can be represent using the expression as follows;

$$\left|\bigcup_{i=1}^n A_i\right| = \sum_{\emptyset\not=N\subseteq\{1,2,\dots,n\}} \left| \bigcap_{r \in N} A_r \setminus \bigcup_{s \in N^\complement} A_s \right|$$

For the cardinality of union of multiple sets, there is inclusion–exclusion principle.

$$\left|\bigcup_{i=1}^n A_i\right| = \sum_{\emptyset\neq J\subseteq\{1,2,\ldots,n\}}(-1)^{|J|-1} \Biggl|\bigcap_{j\in J} A_j\Biggr|$$

From this, we can have

$$\sum_{\emptyset\not=N\subseteq\{1,2,\dots,n\}} \left| \bigcap_{r \in N} A_r \setminus \bigcup_{s \in N^\complement} A_s \right| = \sum_{\emptyset\neq J\subseteq\{1,2,\ldots,n\}}(-1)^{|J|-1} \Biggl|\bigcap_{j\in J} A_j\Biggr|$$ Is this correct?

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Your expression is correct. Given any point $x\in\bigcup_{i=1}^n A_i$, let $N$ be the (necessarily nonempty) set of all $1\le i\le n$ for which $x\in A_i$; then $x$ is in the set $\bigcap_{r\in N} A_r \setminus \bigcup_{s\in N^c} A_s$ for this $$N$ and for none of the others. That proves your disjoint union claim, and the other claims follow quickly from that one.

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