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Suppose that $[n] := \{1,2,...,n\}$. Consider the sets $A_1, A_2, ... , A_n$. Each set $A_i$ is non-empty and a fixed subset of a universal set $\Omega$. I want to create a set containing all the possible intersections of all the sets up to $A_n$. My initial attempt was using

$$\bigcup_{i=1}^{n} \bigcap_{i=1}^{n} A_{i}$$ But I believe that this would simply give me the set equal to the intersection of all the sets, because of how the operator is defined: $\bigcup_{i=1}^{n} A_i = \{ x \in \Omega: \exists i \in \{1,2,...,n\}, \text{such that } x \in A_i$. Which is the same as $A_1 \cup A_2 \cup ... \cup A_n$. The same logic applies to the intersection operator. This would mean that you would only get a set containing all the intersections, but not a set containing all the possible intersections. Therefore my question is, what is the correct way to define such set?.

Take for example $n=3$. Then $\bigcap_{i=1}^{3} A_i = A_1 \cap A_2 \cap A_3$. But this does not take the intersections $A_1 \cap A_3$, $A_2 \cap A_3$ or $A_1 \cap A_2$ into account, only the joint intersection of those sets. It is of course possible that an intersection can be an empty set, and the union can therefore also possibly be an empty set, but this does not have to be the case.

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  • $\begingroup$ If you mean "a set whose elements are everything that is in any of the intersections" then Gribouillis has your answer. If you mean "a set whose elements are the various intersections" then Yozef Tjandra does. $\endgroup$
    – Mark S.
    Commented Jun 12, 2020 at 10:49

2 Answers 2

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Will $$ \left\{ \bigcap\limits_{i\in I} A_i : I\subseteq [n] \right\} $$ work for you?

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  • $\begingroup$ I edited the question slightly. Am I missing something in your reasoning, because I believe that you would just get all the intersections? $\endgroup$ Commented Jun 12, 2020 at 10:09
  • $\begingroup$ @Kevo didn't you want all the intersections? Like if $n=3$ this set has up to $8$ elements (if all the intersections are different) including, for example, $A_1\cap A_3$. $\endgroup$
    – Mark S.
    Commented Jun 12, 2020 at 10:52
  • $\begingroup$ Is this equal to creating a powerset $N = \mathcal{P}([n])$, and then the same logic, but with $i \in N$ $\endgroup$ Commented Jun 12, 2020 at 13:04
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Every $A_j$ is an intersection because $A_j = \bigcap_{i\in\{j\}}A_i$, hence the set you are after must contain all the $A_j$. Finally, the smallest possible set is \begin{equation} S = \bigcup_i A_i \end{equation}

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  • $\begingroup$ I edited the question slightly. I am a bit stuck here. Does your reasoning on intersections not imply that some intersections are left out? $\endgroup$ Commented Jun 12, 2020 at 10:10
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    $\begingroup$ I don't see which intersections are left out. For example $A_1\cap A_2\cap A_3 \subset A_1 \subset S$. $\endgroup$ Commented Jun 12, 2020 at 10:14

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