$E^p$ is a ring, but not a $\sigma$-ring

Question

An interval in $\mathbb{R}^p$ is defined as a set of the form $\{ \vec{x} \in \mathbb{R}^p : \forall i=1,2,...,p \quad a_i \le x_i \le b_i \}$, where the inequalities can be strict or not, and the numbers $a_i,b_i$ are allowed to be infinite. An elementary set is a set which is a union of a finite amount of intervals. The class of all elementary sets is denoted by $E^p$. It is left as a comment in Rudin to verify that $E^p$ is a ring, yet not a $\sigma$-ring, that is a countable union of such sets need not be necessarily elementary. Showing that it's a ring I kind of get (although I don't know if there's a truly simple way of showing that the compliment of an elementary set is elementary), but I'm having trouble thinking up a counterexample to the countable union property...

Answer 1

It's enough to give a counter-example when $p=1$. Consider $A_n:=(n-1/2,n+1/2)$: the union is not elementary.

Answer 2

I think the most intuitive counterexample for this is given by considering connected components. Note that all intervals (your definition) are connected, and that unioning a set with a connected set can at worst increase the number of connected components by 1.

It follows that finite unions of intervals must have finitely many connected components. However, it is easy to construct an example of a set in $\mathbb{R}^n$ which is a union of intervals but has infinitely many connected components, e.g. take integer translates of a small interval around the origin.