Further Useful Partition Results

Let \(A\) be the complement of the closure \(\overline{E}\) and let \(B\) be the complement of the closure \(\overline{E^c}\). Both \(A\) and \(B\) are open sets in \({\mathbb R}^n\) and they are disjoint (since belonging to \(A\) implies not belonging to \(\overline{E}\) and hence not belonging to \(E\) and likewise \(B \subset E^c\)). If \(x \not \in A \cup B\), then necessarily \(x \not \in A\), so \(x \in \overline{E}\) and \(x \not \in B\), so \(x \in \overline{E^c}\). This forces \(x \in \overline{E} \cap \overline{E^c} = \partial E\). If \(R'\) is any connected set (like a box) not intersecting \(E'\), then \(R' = (R' \cap A) \cup (R' \cap B)\), and so necessarily one of these two pieces must be empty. In the former case, it means that \(R' \cap E\) must be empty as well, and in the latter it means that \(R' \cap E^c\) is empty.

For any fixed \(x \in \partial E\), it is always possible to express \(x\) as the limit of a convergent sequence of points \(\{y_n\}_{n=1}^\infty\) in \(E\) and also as the limit of a convergent sequence of points \(\{z_n\}_{n=1}^\infty\) in \(E^c\). Also, because \({\mathbb R}^n = E \cup E^c\), it may be assumed that one of these sequences is constant (take \(y_n := x\) for all \(x\) if \(x \in E\) and take \(z_n := x\) for all \(n\) if \(x \in E^c\)). If \(x\) does not belong to the boundary \(\partial R\) (as otherwise, any box \(R' \in {\mathcal P}\) containing \(x\) also intersects \(\partial R\)), then both sequences must belong to \(R\) as well for all sufficiently large \(n\). Because \({\mathcal P}\) is finite, there must exist a box \(R'\) containing infinitely many terms of the non-constant sequence (e.g. containing \(y_n\) for infinitely many values of \(n\) in the case \(x \in E^c\)). Because \(R'\) is closed and the sequences both converge, \(x\) must belong to \(R\) as well. Thus \(R'\) must contain the point \(x\) and must contain terms from both sequences, meaning it intersects both \(E\) and \(E^c\).

Fix a constant \(c > 1\) to be determined momentarily. For each \(i \in \{1,\ldots,N\}\), let \(c R_i\) be the box with the same center as \(R_i\) with side lengths exactly \(c\) times the side lengths of each \(R_i\). For example, if \(R_i := [a_1,b_1] \times \cdots \times [a_n,b_n]\), then \(c R_i = [\alpha_1,\beta_1] \times \cdots \times [\alpha_n, \beta_n]\) such that \((\alpha_j + \beta_j)/2 = (a_j + b_j)/2\) and \(\beta_j - \alpha_j = c (b_j - a_j)\) for each \(j = 1,\ldots, n\). By design, \(c R_i\) contains \(R_i\) and \(|c R_i| = c^n |R_i|\) for each \(i \in \{1,\ldots,N\}\). Now let \(c\) be any number greater than \(1\) such that \(c^n < 1+\epsilon\). Let \(\tilde R_i := R \cap (c R_i)\) for each \(i\), and let \({\mathcal P}\) be a grid partition of \(R\) which is adapted to \(\{\tilde R_i\}_{i=1}^N\), i.e., the elements of \({\mathcal P}\) which intersect the interior of any \(\tilde R_i\) or \(R_i\) form a partition of it. For each \(i \in \{1,\ldots,N\}\),

since every point of \(R_i\) is contained in the interior of \(c R_i\) and any point in the interior of \(c R_i\) will be in the interior of \(\tilde R_i\) as well unless it falls outside \(R\) (which never happens for points in \(R_i\)) or lies on the boundary of \(R\). So any box \(R''\) intersecting an \(R_i\) also intersects the interior of \(\tilde R_i\) or its common boundary with \(R\). In either case, it must be contained in \(\tilde R_i\). To conclude,