[Math] Boundary of a countable union sets contained in the union of boundaries

Question

First off, this is related to my previous question. Suppose that I have a countable collection of mutually disjoint open cubes $\{Q_k : k \in \mathbf{Z}^+\}$ in $\mathbf{R}^n$. What can I say about the boundary of the union? That is, what can I say about the boundary of
$$
Q = \bigcup_{k = 1}^\infty Q_k?
$$
What I really want is that the boundary of $Q$ is contained in the union of the boundaries of the cubes. In general topology, you cannot say that the boundary of the union is contained in the union of the boundaries. For instance, enumerate the rationals in $(0,1)$ by $R = \{r_k\}$. Then the boundary of the union is $[0,1]$, but the union of the boundaries is $R$ itself. I think that in this less general situation of cubes, it may be the case that the boundary of the union is contained in the boundary of the unions, but I have not been able to prove it. Any ideas are appreciated! Thanks!!

Answer 1

Here's a counterexample: let $n=1$ and $Q_k=(\frac1{k+1},\frac1k)$. Then the union of boundaries is $$\left\{\frac1k\Bigg|\;k\in\mathbb N\right\}$$ and the boundary of the union is $$\left\{\frac1k\Bigg|\;k\in\mathbb N\right\}\cup\{0\}.$$