Conditions so that Lebesgue Covering Dimension and "Usual" Dimension are Equal The definition of covering dimension is as follows: The ply of a cover is the smallest number $n$ (if it exists) such that each point of the space belongs to at most n sets in the cover. A refinement of a cover $C$ is another cover, each of whose sets is a subset of a set in $C$; its ply may be smaller than, or possibly larger than, the ply of $C$. The covering dimension of a topological space $X$ is defined to be the minimum value of $n$, such that every finite open cover $C$ of $X$ has a refinement with ply at most $n + 1$. If no such minimal n exists, the space is said to be of infinite covering dimension.
Dimension is also a concept tossed around in linear algebra to describe the number of basis elements which are necessary to generate the space in question. I feel that these two notions should coincide with a few conditions on the space in question - I have already been able to show that plane figures like squares have covering dimension 2 etc. I think it's reasonably clear that metrizability is a requirement, but I can't write down a rigorous argument based only on this.
Can anyone give me any pointers on how to come up with the other necessary and sufficient conditions?
 A: The idea of the covering dimension is due to Lebesgue, who claimed to have proven that $[0,1]^n$ has covering dimension $n$ (in modern terms; he formulated it in terms of properties of finite covers of sets of small diameter; Cech later gave a more formal definition of dimension based on the order (or ply) of a cover). Brouwer, somewhat later actually gave a valid proof and proposed a dimension function defined by induction. He also shows that under his inductive dimension definition (he called it Dimensionsgrad in German) the dimension of $\mathbb{R}^n$ was $n$ (solving the up to then open problem of whether $\mathbb{R}^n$ and $\mathbb{R}^m$ were homeomorphic iff $n=m$). So this started the whole field of topological dimension theory, see for more details this encyclopedia article. The proof that $\dim(\mathbb{R}^n) = n$ topologically is actually quite tricky.
The linear (linear algebraic) dimension of $\mathbb{R}^n$ is of course $n$, as was well-understood at the time, and this corresponds well to the "$n$ coordinates to describe a point" idea of dimension. Of course space filling curves by Peano and bijections between $\mathbb{R}^n$ and $\mathbb{R}^m$ by Cantor formed the whole reason to give a more topological, intrinsic definition of dimension.
Of course some discrepancies exist: $\mathbb{C}^n$ is $n$-dimensional over the field $\mathbb{C}$ but $2n$-dimensional over $\mathbb{R}$; topolically it has dimension $2n$ of course (and infinite dimensional over $\mathbb{Q}$). So linear dimension presupposes a field over which this dimension is measured.
It turns out that the inductive dimensions (there are two, see the article referred to above) and the covering dimensions coincide for all separable metrisable spaces (so there really is one notion of topological dimension for that class of spaces), but for spaces outside of that class, differences can occur. 
When both linear dimension (over the reals (!)) and topological dimension apply and are finite, we are essentially (topologically) working in $\mathbb{R}^k$ and so the definitions coincide. If a vector space is infinite-dimensional over the reals in a linear sense, it is also topologically infinite-dimensional. So we have agreement there as well.
