How to calculate topological dimension of $\mathbb{R}^n$ I'm reading Engelking - Dimension Theory to verify that $\mathbb{R}^n$ has topological dimension $n$. (Sometimes called covering dimension)
In this book, the author uses small and large inductive dimensions, ind e Ind, explores the relation among theses three dimensions. To do so, he needs a lot of results.
Can you indicate a quicker proof without algebraic topology, please?
 A: The proof in Engelking does not use algebraic topology, it's a purely topological proof.
However, Engelking is writing a general text, and strives for as large a generality as can be achieved easily. This means that there are a lot of intermediate results, and general results on all three main dimension functions $\operatorname{ind},\operatorname{Ind}$ and $\dim$. The fact that $\dim(\Bbb R^n) = n$ is not proved in the most direct manner, indeed.
In his book "General topology" he does a shorter proof, but this proof assumes a lot of the general material from earlier in his book as well. It is somewhat more to the point though. Engelking's "Topology: a geometric approach", also features simplices and Brouwer's theorem (and does mostly metric spaces), but also introduces homotopy groups. So I'm not sure if this would qualify for your conditions.
I know of a purely metric approach in Jan van Mill's "Infinite dimensional topology, prerequisites and an introduction" (where dimension theory is one chapter of prerequisites). One also has to prove the Brouwer fixed point theorem, which is needed in the proof (also in Engelking's one). And this is done via Sperner's lemma which is a combinatorial theorem on simplicial complexes (so these need to be developed also). Jan van Mill's later book ("the infinite dimensional topology of function spaces") also features these proofs.
A: I commented about algebraic topology because there are proofs with such approach and it's not interesting for me, by now.
The thing about Engelking - General Topology and Engelking is that his definition of topological dimension is
"a normal space $X$ has topological dimension $n$ if $n$ is the smallest natural such that every finite open cover of $X$ has a finite open refinement of order $\leqslant n$"
The definition I work with doesn't requires finiteness and until now I can't understand why I could resctrict to finite open cover. The space is not compact necessarily.
I'm gonna check this book you cited.
Thanks.
