How can one show that the topological dimension (Lebesgue covering dimension) of the closed $n$-dimensional unit ball $\{x\in\mathbb{R}^n\mid 1\geq|x|\}$ is indeed $n$, assuming only that:
- the topological dimension of $\mathbb{R}^n$ equals $n$.
- the topological dimension of open $n$-dimensional unit ball $\{x\in\mathbb{R}^n\mid 1>|x|\}$ equals $n$.
- the topological dimension is invariant under homeomorphism.
I know that this can be shown using the small and large inductive dimensions $ind$ and $Ind$. But is there a way that minimises the need for theoretical background knowledge and only uses above assumptions?
My approach was to show that each finite open cover and its refinement of the open unit ball could be extended by a tiny enlargement to a cover and its refinement of the closed ball and thus show that the dimensions of open and closed balls are equal. However, I think that this fails because not all finite open covers of the open ball have an equivalent for the closed ball.
Edit: Additional assumption:
- Let $A$ be a closed subset of the metric space $S$. Then $\dim(A)\leq\dim(S)$