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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!!

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Suugaku
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2 Answers2

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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 $$\left\{\frac1k\Bigg|\;k\in\mathbb N\right\}\cup\{0\}.$$

Son Gohan
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Dejan Govc
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  • Thank you! This definitely is a good counterexample. Now let's take this question even further. What I really need is $\overline{Q}\backslash \cup Q_k$ to be of measure zero. Here $\overline{Q}$ denotes the closure of the union of all the cubes. Of course if what I posted were true, then this would but true. As your example shows, it's not true that the boundary of the union is contained in the union of the boundaries. However, in your example given, it is indeed true that the difference has zero measure. Is it always true? – Suugaku Apr 27 '13 at 03:48
  • @Suugaku: Not necessarily: https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set – Dejan Govc Apr 27 '13 at 11:28
  • @Suugaku: For a counterexample in higher dimensions, a "fat" version of the Sierpiński carpet should do the job. To get positive results, I suspect something like local finiteness of the collection of cubes might help. – Dejan Govc Apr 27 '13 at 11:42
  • Thanks again! I figured a good counterexample would come out of the Cantor set. I didn't think to look at the fat one. As far as the local finiteness thing goes, the collection I'm looking at is the set of open, mutually disjoint, dyadic cubes obtained from the Calderon-Zygmund decomposition. I'm not sure if these cubes are "nicer" or if they can in fact be formed like the fat cantor set? – Suugaku Apr 27 '13 at 17:41
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It's not true. In $\mathbb{R}^1$, consider the set of one intervals $\{(1/n,1/(n+1)\}$. The point 0 is in the boundary of the union, but no the union of the boundaries, and these are disjoint open cubes.

Brian Rushton
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