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What does it mean by pairwise intersection? Is it the same as finite intersection?

First, it is trivial that if a set is closed under finite intersection, it is closed under pairwise as well. Now, I show a set closed under pairwise intersection is also closed under finite intersection.

Suppose $\mathcal{S}$ is closed under pairwise intersection. Let $A, B, C$ be given such that they are in $\mathcal{S}$. Then $A\cap B\in\mathcal{S}$. Since $C\in\mathcal{S}$, $(A\cap B)\cap C\in\mathcal{S}$. Induction gives us that any finite intersection is also in $\mathcal{S}$.

So it seems like they are really the same concept, right?

Asaf Karagila
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Kun
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    I tried to google, no luck... – Kun Mar 21 '16 at 15:02
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    Your reasoning is perfectly valid: indeed these two concepts are equivalent. This is why you don't find anything about "pairwise intersection": it's a non-standard way of saying "finite intersection". – Crostul Mar 21 '16 at 15:07
  • @Crostul Would you please post it as an answer please? – Kun Mar 21 '16 at 15:08
  • Yes, they're the same: if a set is closed under an binary operation pairwise, then it's closed under all finite combinations under that operation. – BrianO Mar 21 '16 at 17:41

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The operations are not equivalent, because closure under empty intersections does not follow from closure under pairwise intersections. However, an empty intersection is clearly a finite intersection.

$\pi$-systems are families of sets which are closed under pairwise intersections (hence all non-empty finite intersections). However, they do not have to be closed under empty intersections, hence do not have to be closed under all finite intersections.

Your induction proof only works to show closure under non-empty finite intersections. In particular, it's not possible to induct "backwards" from $n=1,2$ to $n=0$ (empty intersection).

The question and answers on this page explain empty intersection better than I can.

See also this question.

Chill2Macht
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  • I know this seems like obnoxious pedantry, but the distinction is emphasized heavily, for example, in the definition of lattice: https://en.wikipedia.org/wiki/Completeness_(order_theory)#Finite_completeness Namely a lattice has to be closed under non-empty finite meets and joins, but not all lattices are bounded, i.e. also closed under the empty meet and the empty join. – Chill2Macht Dec 19 '17 at 20:20