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Suppose there are events $E_i$ for integer values such that $1\leq i\leq k$. Now let some set of sets (call it $G$) be composed of every set that is the intersection of all $k$ events or (exclusive or) their complements. I'm finding it difficult to word this correctly, so this should hopefully help:

For example, if $k=2$, then $G$ should ONLY contain:

${E_1 \cap E_2}$,

${E_1 \cap E_2^{'}}$,

${E_1^{'} \cap E_2}$,

${E_1^{'} \cap E_2^{'}}$

Where the apostrophe denotes the event's complement. Is there a name for such a set $G$?

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3 Answers 3

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We could call it the partition generated by the class $\{E_i\}_{i=1}^k$.

Some examples of this usage:

  • Dorian Feldman and Martin Fox, Probability: The Mathematics of Uncertainty, Section 4.B: Partition Generated by a Class
  • Paul Pfeiffer, Concepts of Probability Theory, Definition 2-7b
  • Achim Klenke, Probability Theory, proof of Hahn's decomposition theorem:

    Define $A_n^0 := A_n$, $A_n^1 := A \setminus A_n$, and let $$\mathcal P_n := \left\{ \bigcap_{i=1}^n A_i^{s(i)} : s \in \{0,1\}^n\right\}$$ be the partition of $A$ that is generated by $A_1, \dotsc, A_n$.

  • Math.SE: Partition generated by a class of subsets
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The collection G of sets is closed under intersections and complements.

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    $\begingroup$ Is that entirely correct, though? For example, it doesn't contain $E_1 \cap E_1^{'}$ Also, suppose my k was larger, such as 3. Then $G$ would contain $E_1 \cap E_2 \cap E_3$ but not $E_1 \cap E_2$. $\endgroup$ Commented Nov 17, 2017 at 5:30
  • $\begingroup$ @JohnTravolski It does contain $E_1 \cap E_1'$ - see my answer. Either that or you have to rephrase more carefully what your system is supposed to be. $\endgroup$ Commented Nov 17, 2017 at 5:35
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Observe that your system is closed under complementation since for a given $E$ in your system, $E' = E' \cap E'$ is also in your system.

Since your system is closed under complementation and finite intersection, it's also closed under finite union. Why? Let $E,F$ be in your system. Then $$ E \cup F = (E' \cap F')' $$ is in your system as well.

Unless your system is empty, it contains the emptyset (since it is closed under complementation and intersection) and thus is also contains $A$. This sort of system is known as an algebra and well studied in mathematics. (Compare to $\sigma$-algebra but we don't require algebras to be closed under countable unions or intersections.) .

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