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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.

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  • $\begingroup$ 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? $\endgroup$
    – BSteinhurst
    Commented Apr 7, 2011 at 16:55
  • $\begingroup$ 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? $\endgroup$
    – Valerio Capraro
    Commented Apr 7, 2011 at 17:17
  • $\begingroup$ 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. $\endgroup$
    – BSteinhurst
    Commented Apr 7, 2011 at 18:23

1 Answer 1

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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.

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  • $\begingroup$ 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 $C=d(x,y)$. $\endgroup$
    – Valerio Capraro
    Commented Apr 8, 2011 at 11:55
  • $\begingroup$ It is not even true that I suggested to "perform a suicide" :). I add few lines, hope it helps. $\endgroup$
    – Anton Petrunin
    Commented Apr 8, 2011 at 16:29
  • $\begingroup$ Wait a moment. Let's talk a bit more specifically. As I said at the end of my post, actually the context is more general. It's the context of convex-like structures in the sense of Nate Brown (you might know him, being at Penn State) and we don't know at the moment if they are isometrically and linearly embeddable into a metric vector space. But anyway your remark is very important: the metric is additive on lines. What do we know? We know that a convex-like structure embeds linearly into a vector space. So we basically have a vector space and a convex subset equipped with a metric which is $\endgroup$
    – Valerio Capraro
    Commented Apr 9, 2011 at 14:32
  • $\begingroup$ additive on lines. The question is: Is always possible to extend the metric to the whole space in such a way to get a norm space? If the answer is positive, it would be great! If not, let's go back to the initial question: is it possible to prove that finite dimensional (in the sense of the simplex-dimension) convex-like structure are bi-lipshitz to euclidean spaces? Too many questions?? $\endgroup$
    – Valerio Capraro
    Commented Apr 9, 2011 at 14:37
  • $\begingroup$ I mean: in case can you give me a sketch of a reference of the proof that finite-dimensional subspaces of a normed space are bi-lipschitz to euclidean space, so that I can understand if it is adaptable. $\endgroup$
    – Valerio Capraro
    Commented Apr 9, 2011 at 14:42

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