# Closed under intersections

I read this definition: "A collection C of subsets of E is said to be closed under intersections if A ∩ B belongs to C whenever A and B belong to C."

How could the intersection of ANY A and B belonging to C ever NOT belong to C?? Whats the point of this definition?

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Let $E=\{1, 2, 3\}$ and suppose the collection $C$ of subsets was $C=\{\{1, 2\}, \{2, 3\}\}$. Then this collection of two subsets is not closed under intersection, since $\{1, 2\}\cap\{2, 3\}=\{2\}$, which is not in $C$.

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Oh, now I think I get it..."collection" is the key word here!? –  PrincessAmazia Oct 10 '13 at 1:14
Yup. Your collection of subsets is pretty much arbitrary, so the intersection of two of the subsets needn't be a set in the collection. Of course, any 1-subset collection will be closed under intersection, but beyond that there's no guarantee of closure. –  Rick Decker Oct 10 '13 at 1:16

Let $A=\{a,b\}$ and $B=\{b,c\}$. Define $\mathscr{C}=\{A,B\}$. Then $A,B\in\mathscr{C}$, but $A\cap B\notin\mathscr{C}$.

The point of this definition is to enforce some restrictions on $\mathscr{C}$ so you can have 'operations' on some of the elements (like intersections).

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Very easily. Let $C=\{(0,2),(1,3)\}$, where $(0,2)$ and $(1,3)$ are subsets of $\Bbb R$. The intersections of members of $C$ are

$$(0,2)\cap(0,2)=(0,2)$$

and

$$(1,3)\cap(1,3)=(1,3)\;,$$

which are in $C$, and

$$(0,2)\cap(1,3)=(1,2)\;,$$

which is not in $C$.

It can happen just as well with infinite collections. For each $n\in\Bbb N$ let $A_n=\{n,n+1\}$, and let $\mathscr{C}=\{A_n:n\in\Bbb N\}$; the members of $\mathscr{C}$ are the sets $\{0,1\},\{1,2\},\{2,3\},\ldots\;$. For any $m,n\in\Bbb N$ we have

$$A_m\cap A_n=\begin{cases} A_m,&\text{if }m=n\\ \{n\},&\text{if }n=m+1\\ \{m\},&\text{if }m=n+1\\ \varnothing,&\text{in all other cases}\;. \end{cases}$$

Only in the first case is the intersection in the collection $\mathscr{C}$.

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