Q:

Proving the set of all finite or countable unions of intervals is not a Sigma Algebra

I would like to extend on a question I asked here 
Consider a set $J$ of all (open, closed, half-open, singleton, empty) intervals on $[0,1]$
Now consider further a set $B$ which is the set of all finite or countable unions of elements of $J$. 
According to the text I'm reading, $B$ is not a $\sigma$-algebra.
I suspect that it is because it is not closed under countable intersections, however I can't understand why. Surely any countable intersection is simply an interval or the empty set? Can't come up with any sort of contradiction. 

A:

The complement of the middle-thirds Cantor set is a countable union of (open) intervals.
However, the Cantor set itself has uncountably many elements, and any two of them are separated by a point not in the Cantor set. So no countable union of intervals can produce it.
Thus $B$ is not closed under complement and therefore it is not a $\sigma$-algebra.

