Boundary of an open set = Boundary of the connected components? Suppose $\mathcal{O} \subseteq \mathbb{R}^n$ is open. Then we can write $\mathcal{O}= \bigcup_{i \in I} \mathcal{O}_i$, where $\mathcal{O}_i$ are the connected components of $\mathcal{O}$. Now I wonder if we have 
$$\bigcup_{i \in I} \partial \mathcal{O}_i  \overset{?}{=} \partial \mathcal{O}.$$ 
If there is an example where $\bigcup_{i \in I} \partial \mathcal{O}_i  \neq \partial \mathcal{O}$, the question is, if we can say that either "$\subseteq$" or "$\supseteq$" does hold in general.
 A: Consider the Sierpinski gasket. Consider the open set $O$ obtained by taking the interior of all the triangles appearing in Sierpinski's gasket. The connected components $O_n$'s are just the interiors of the triangles. So the union of boundaries $\partial O_n$ is just the union of the triangles which is the Sierpinski gasket. But since $O$ is dense in big triangle, there are continuum many points belonging to the boundary of $O$ which do not lie on any of the triangles $\partial O_n$ - Just consider a line perpendicular to the base of the big triangle which does not pass through any of the vertices of the countably many triangles in the gasket. Each triangle that meets this line meets it at a closed interval and these intervals are pairwise disjoint so there are continuum many points on this line which do not belong to $\bigcup_n (U_n \bigcup \partial U_n)$.
Also, $\bigcup_n \partial O_n \subseteq \partial O$ is true because every boundary point of $O_n$ is not in $O$ so it is a boundary point of $O$.
A: Inclusion "$\subseteq$" is true. Opposite inclusion is false. Counterexample in $\mathbb{R}$ is sum of intervals:
$$\bigcup_{k\in \mathbb{N}_+} \left(\frac1{k+1},\frac1{k}\right).$$
The point $0$ is included in right side but not in left side.
