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.
A: Here is a formal argument which explains and simplifies the argument given by hmakholm left over Monica. Introducing a Cantor set is not necessary.
I call $\mathcal{J}$ the set of all finite (possibly empty) or countable union of intervals. If $B \in \mathcal{J}$, then $B$ has at most countably many components.
Indeed, write $B$ as the union of non-empty intervals $I_n$, where $n$ varies in some finite or countable set $S$. Choose a point $a_n$ in each interval $I_n$, and call $C(a_n)$ its connected component in $B$. Since all points of $I_n$ have the same connected component in $B$, the map $n \mapsto C(a_n)$ is a surjection from $S$ to the set of all connected component in $B$.
The set $\mathbb{Q}$ is a countable union of single sets, hence belongs to $\mathcal{J}$. However, $\mathbb{R} \setminus \mathbb{Q}$ does not belong to $\mathcal{J}$ since it has uncountably many connected components: its connected components are the single sets $\{x\}$ for $x \in \mathbb{R} \setminus \mathbb{Q}$.
