Calculating the boundary of a specific subset of real numbers We construct a sequence of open sets $J_i$as follows. Let $J_0$ be the set $\cup_{n\in \mathbb Z} (2n,2n+1)$. Having constructed $J_0,J_1,...,J_{m-1}$ let $J_m$ be the union of the open middle thirds of the segments constituting $\mathbb R - \cup_{i=0}^{m-1} J_i$. ( I am not sure if the following is an algebraic formulation although as far as I remember I obtained it right: Consider the subsets {$J_m| m=0,1,2,...$} of $\mathbb R$ defined by $$J_m:= \bigcup_{s=0}^{m-1} \bigcup_{n\in \mathbb Z} \left(\frac{3^m(2n+1)+2\times 3^{s}+1}{3^m},\frac{3^m(2n+1)+2\times 3^{s}+2}{3^m}\right),$$ where by definition $2\times 3^0:=0$. If this algebraic formulation is either wrong or right but not useful, dismiss it.)
1)  Is it true that the boundary of the union of any infinite subcollection of {$J_m$} equals the same set (maybe the set of all endpoints of the intervals constituting $J_m$’s)?
2)  If not, according to my book from which this question has arisen why the boundary of $\cup_k J_{3k}$, the boundary of $\cup_k J_{3k+1}$, and the boundary of $\cup_k J_{3k+2}$ are the same?
 A: At every step, $\mathbb{R}-\cup_{i=0}^{m} J_i$ consists of a bunch of disjoint intervals of length $3^{-m}$.  This is true for $m=0$ by construction, and inductively at each stage the middle third of each interval in $\mathbb{R}-\cup_{i=0}^{m} J_i$ becomes part of $J_{m+1}$, leaving two intervals with a third the length.  
Every boundary point of $J_l$ is a boundary point of $\mathbb{R}-\cup_{i=0}^{m} J_i$ as long as $l\leq m$, and every boundary point of $\mathbb{R}-\cup_{i=0}^{m} J_i$ is within a distance $3^{-m}$ of some boundary point of $J_{m+1}$.  Hence, given any infinite collection of $\{J_i\}$, we can construct a sequence of boundary points $\{x_i\in J_i\}$ from them which converges to any boundary point of $J_l$.  
Since the $J_i$ are open and disjoint, their boundary points are not contained in any other $J_i$'s.  From this it follows that the boundary of the union of a collection $\{J_i\}$ contains the union of their boundaries.  For any boundary point $x$ of some $J_l$ is not in $\cup \{J_i\}$, but it is arbitrarily close to points thereof, hence it is in the boundary of $\cup \{J_i\}$.
Hence, given any infinite collection $\{J_i\}$, we can find a sequence in the boundary of their union converging to any boundary point of any $J_l$.  Since the boundary is closed, this means that any boundary point of $J_l$ is in the boundary of the union of $\{J_i\}$.  This proves your point $1$.
