# Lebesgue covering dimension

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.

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.

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If you are using the simplex-dimension to prove theorems, why do you care what the covering dimension is? Or are you in a situation where you know the covering dimension and want to equate it to the simplex-dimension? –  BSteinhurst Apr 7 '11 at 16:55
Is it called simplex-dimension? Because... no particular reason, just curiosity and, why not, another theorem. It's a metric space, then normal, then, at least in the separable case, all the inductive dimensions and the covering one coincide. I suspect they also coincide with the simplex-dimension. What about the Hausdorff dimension? –  Valerio Capraro Apr 7 '11 at 17:17
There are so many notions of dimension that it is useful to have a way to distinguish them, so I've never heard "simplex-dimension" before but it communicated to you what I wanted it to so that is some use. I have a feeling that in convex spaces these notions should be the same. The usual double inequality style of argument should probably do the trick. –  BSteinhurst Apr 7 '11 at 18:23

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.
I hope you are not going to call me "stupid", as written in your profile, but I don't understand why (1) should imply the equality in (2) with $C=d(x,y)$. –  Valerio Capraro Apr 8 '11 at 11:55