# $\mathrm{dim}(\partial U_k)\leq n$ for all $k$, implies $\mathrm{dim}(\partial (\bigcap_k U_k))\leq n$?

Let $$X$$ be a locally compact, Hausdorff, second countable space. Let $$(U_k)_{k\in\mathbb{N}}$$ be a countable family of closed subsets in $$X$$ with $$\mathrm{dim}(\partial_X U_k)\leq n$$, for all $$k$$ (where $$\mathrm{dim}$$ stands for topological covering dimension). Does it follow that $$\mathrm{dim}(\partial_X (\bigcap_{k\in\mathbb{N}} U_k))\leq n ?$$

If the collection was finite, then the answer would be 'yes', as $$\partial_X (\bigcap\limits_{k=1}^{m} U_k)\subseteq \bigcup\limits_{k=1}^{m} \partial_X U_k$$.

It is not clear that a similar inclusion holds for infinite collections, as here Boundary, unions and intersections I think that there is a mistake in the answer, the inclusion $$\mathrm{int}( \bigcap \limits_{A \in \mathscr{A}} A)\subseteq \bigcap \limits_{A \in \mathscr{A}} \mathrm{int}(A)$$ is reversed.

If the answer is 'no', I would be still happy to know about a class of topological spaces in which this holds (namely, having more restrictions on $$X$$).