Roughly from wikipedia: The covering dimension of a topological space $X$ is defined to be the minimum value of $n$ such that every finite open cover of $X$ has a finite open refinement in which no point is included in more than n+1 elements. If no such minimal n exists, the space is said to be of infinite covering dimension.
Now, I have to say that I don't get on with this definition. In particular, my situation is the following. I have a vector space $V$ which is complete with respect to the metric $d$. Moreover the linear structure and the metric agrees enough well, in the sense that the following are true
1) $d(tx+(1-t)y,tz+(1-t)w)\leq td(x,z)+(1-t)d(y,w)$ for all $x,y,z,w\in V, t\in[0,1]$
2) $d(sx+(1-s)y,tx+(1-t)y)\leq C|t-s|$, where $C$ is a constant that depends on $d(x,y)$ and converges to zero whenever $x\rightarrow y$.
Analogue properties hold for convex combinations of $n$ vectors.
In this context it seemed to me quite natural to define the dimension of $V$ as the greatest $n$ such that any ball of $V$ contains an homeomorphic copy of an $n$-dimensional simplex [see the end of the post].
My question is: Is this dimension equal to the Lebesgue one?
En passant I would be also interested in the compactness of such spaces. It's indeed quite easy to construct examples of bounded metric linear space with dimension =1 (in the sense of the greatest $n$ such that there is a $n$-dimensional simplex) that are not compact, but they do not satisfy the additional properties I wrote above.
Thanks in advance, Valerio
One could ask whether I haven't choosen an easier definition for the dimension. There are basically three reasons: the first one is that actually I don't know if my space is into a vector space, but at the moment it just verifies the axioms of convex space in the sense of Marshall Stone. The second reason is that I want some isotropy of the space: I want that the inside the balls I can find more or less the same things. The third is the most natural one: this is the definition that allows to prove some nice result.
Best Answer
If you take in (1) $x=z=w$ and then you get $$d(tx+(1-t)y,x)\leq (1-t){\cdot}d(x,y),$$ the same way you get $$d(tx+(1-t)y,y)\leq t {\cdot}d(x,y).$$ By the triangle inequality, you have "=" in the both of these inequalities. I.e., $$d(tx+(1-t)y,x)= (1-t){\cdot}d(x,y).$$ Applying it twice, you get equality in (2) with $C=d(x,y)$. I.e., your metric is induced by a norm.
So any finite dimensional subspace is bi-Lipschitz to Euclidean space. Hence you get that virtually all dimensions of your space coincide.