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?
 A: 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 $C$.
A: 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).
A: 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 $\mathscr{C}$.
