How many open sets with common boundary could be on $\mathbb{R}$? Today I tried to find how many open sets with common boundary could be on $\mathbb{R}$. Actually for $n = 1$ it's obviously. But what about more open sets.
I thought about Cantor set. Because we divide it into a lot of open sets. But I don't know about boundary of this sets. That's the main problem for me. Any hints? 
 A: Look at $n$ points $x_1<...<x_n$ lying in $\Bbb R$ and then the intervals $I_i:=(x_i,x_{i+1})$. Now the $n$ sets
$$A_j:=\Bbb R-\left(\bigcup_{i=1}^n\{x_i\}\cup I_j\right)$$
are all different and their boundaries are all the same, namely $\{x_1,..,x_n\}$.
A: disjoint case
Consider the construction of the middle-thirds Cantor set $C$.  Let $V_n$ be the open set removed at stage $n$.  So
\begin{align}
V_1 &= \left(\frac{1}{3},\frac{2}{3}\right)
\\
V_2 &= \left(\frac{1}{9},\frac{2}{9}\right) \cup
\left(\frac{7}{9},\frac{8}{9}\right)
\\V_3 &= \left(\frac{1}{27},\frac{2}{27}\right) \cup
\left(\frac{7}{27},\frac{8}{27}\right) \cup
\left(\frac{19}{27},\frac{20}{27}\right) \cup
\left(\frac{25}{27},\frac{26}{27}\right)
\\
&\quad\dots
\end{align}
Now for any sequence $n_1 < n_2 < \cdots$ of positive integers, I claim
$$
\bigcup_{k=1}^\infty V_{n_k}
$$
has boundary $C$.  And disjoint sequences will correspond to disjoint open sets.  Say $2^k$ for the first sequence, $3^k$ for the second squence, $5^k$ for the third sequence, and so on with powers of the $j$th prime for the $j$th sequence.
A: Essentially based on @s.harp's answer, every set $A_I := ⋃_{i ∈ ℤ} (2k, 2k + 1) ∪ ⋃_{i ∈ I} (2k + 1, 2k + 2)$ for $I ⊆ ℤ$ has $ℤ$ as boundary. That gives $2^ω$-many different open sets with the same boundary, and this is maximal possible cardinality since there are $2^ω$ open sets at total.
@GEdgar's answer gives $ω$ disjoint open sets with the same boundary, which is again maximal possible cardinality since $ℝ$ is ccc.
