$\dim_{cov}(X)=\max\{\dim_{cov}(K), \dim_{cov}(X\backslash K)\}$ for $X$ compact metric space, $K$ compact?

Question

The question is described the title, let me write it again here:

Let $X$ be a compact metric space and let $K$ be a closed subset of $X$.

Does it follow that the covering dimension of $X$ is the maximum of the covering dimensions of $K$ and $X\backslash K$?

For metric spaces we have $\dim(M)\leq \dim(X)$ for any subspace $M$ of $X$.Thus the inequality $\dim(X)\geq\max\{\dim(K),\dim(X\setminus K)\}$ is clear. I don't know why and whether the converse is true. I would be happy to a reference or counter example.

Thanks

Answer 1

Corollary 1.5.4 in the classic "Theory of dimensions, finite and infinite" by Engelking is the following theorem, a corollary to the sum theorem 1.5.3 above it:

If $X$ is separable metric and $X = \cup_{i=1}^\infty A_i$ where for all subspaces $\operatorname{ind}(A_i) \le n$ and all $A_i$ are $F_\sigma$ sets, then $\operatorname{ind}(X) \le n$ as well.

From this a positve answer to your question immediately follows (knowing some extra things):

Taking $A_1 = K$ and $A_2=X\setminus K$, which are closed (trivialy $F_\sigma$) resp. open (so $F_\sigma$ in any metric space), and if you like we can take $A_i = \emptyset$ for $i \ge 3$. Also, $X$ compact metric implies $X$ separable, and $\operatorname{ind}(X) = \dim(X)$ (the coincidence theorem) for $X$ separable metric, and we can take $n= \max(\dim(K), \dim(X\setminus K))$.