An exercise in Munkres' Topology (Exercise 50.8) shows that if $X$ is a $\sigma$-compact Hausdorff(*) space s.t. every compact subset has topological dimension $\leq n$, then so does $X$. If we define $U(\epsilon) = \{x\in\Bbb{R}^n : B(x,\epsilon)\subset U\}$, then $U(\epsilon)$ is compact $\forall\epsilon>0$, $U = \bigcup_{n=1}^{\infty} U(\frac{1}{n})$, and $U(2\epsilon)\subset\mathrm{Int}(U(\epsilon))\,\,\forall\,\epsilon>0$, so that $U$ has dimension $\leq n$, and it is easy to then show that $U$ has dimension $n$.
My question is whether $\partial U$ must have dimension $n-1$.
Since $\partial U$ is closed, it must have dimension $\leq n$, and for the trivial cases, such as when $U$ is a sphere or a cube, it is easy to see that $\partial U$ is a compact topological manifold, with each point having a neighborhood homeomorphic to $\Bbb{R}^{n-1}$, and by the exercise above, we can see that $\partial U$ must have dimension $n-1$. Clearly (by the notes below), $\partial U$ must have dimension $\leq n$. But is this upper bound strict? Are there examples where $\partial U$ has dimension $\leq n-2$? What if we relax the condition of boundedness?
Definitions:
Let $X$ be a space. If $\mathscr{A}$ is an open cover for X, then $\mathrm{sup} \{ \# (\{ A\in\mathscr{A} \mid x\in A\} ) \mid x\in X\}$, the maximum number of sets intersecting at any one point, is called the order of $\mathscr{A}$. If $\exists m<\infty$ s.t. any open cover $\mathscr{A}$ has a refinement of order $\leq m+1$ that is also an open cover for $X$, then X is said to be finite-dimensional. The dimension of $X$ is defined to be the minimal such $m$. It is a not-so-difficult exercise to prove that any compact subset of $\Bbb{R}^m$ has dimension $\leq m$, and has dimension $m$ if it has nonempty interior. Another fact, relatively easy to show, is that if $X$ has dimension m and $C\subset X$ is closed, then $C$ has dimension $\leq m$.
A space $X$ is $\sigma$-compact if $\exists$ a countable collection $\{X_n\}$ of compact subsets of $X$ s.t. $X = \bigcup_n \mathrm{Int}(X_n)$. For example, $\Bbb{R}^n$ is $\sigma$-compact under the standard topology $\forall n\in\Bbb{N}$.
A corollary of the exercise is that $\Bbb{R}^n$ has topological dimension $n$.
The topological dimension of $\partial U$ is exactly $n-1$.
A remark on definition: by a theorem of Urysohn, the covering definition of dimension agrees with inductive dimensions for all separable metrizable spaces; in particular for subsets of $\mathbb{R}^n$.
Since $U$ is open, $\partial U$ has empty interior; therefore its topological dimension is at most $n-1$.
On the other hand, if there existed a nonempty bounded open set with topological dimension $\le n-2$, then by shrinking and translating it, we would have a basis of such sets for $\mathbb{R}^n$. The definition of small inductive dimension then implies that $\mathbb{R}^n$ has dimension $\le n-1$, which contradicts the well-known fact that this dimension is $n$.