Let $X$ be a compact Hausdorff space whose Lebesgue covering dimension is at least 1, and let $F\subset X$ be a closed subset. We know that $\dim(F)\leq \dim(X)$.
Suppose we assume that $X\setminus F$ is dense in $X$, then can we conclude that $$ \dim(F)\leq \dim(X)-1? $$ I started with the proof that $\dim(F)\leq \dim(X)$: Take a finite open cover $\{U_i\}$ of $F$, then $\{U_i\}\cup\{X\setminus F\}$ is an open cover of $X$. This has a refinement whose order is $\leq n+1$. Intersecting the members of the refinement with $F$ gives me a refinement of $\{U_i\}$ that has order $\leq n+1$.
Given that $X\setminus F$ is dense, I want to then conclude that this refinement has order $\leq n$, but I am not sure how to do this, and I am not even sure if this statement is true.
Any suggestions would be much appreciated. Thanks.
Edit: It appears that the result is true if we assume that $X$ is a manifold (See Corollary 20 from this link). However, I am not sure about the general case.
Not necessarily. For instance, let $X$ be a Cartesian product of a convergent sequence $\{0\}\cup\{1/n: n\in\Bbb N\}$ with a unit segment $[0,1]$, and $F=\{0\}\times [0,1]$.