Let $n\in\mathbb N$ and $X$ be a complete metric space.



> 
> Assume that there is $\epsilon>0$ such that 
>  $$\dim B\_\epsilon(x)\le n$$
>  for any $x\in X$.
>  Is it true that $\dim X\le n$?
> 
> 
> 


* Here $\dim$ stays for [topological dimension](http://en.wikipedia.org/wiki/Lebesgue_covering_dimension).
* We do not assume that $X$ is separable!!!



Topological dimension, is it local?
From [E]T.7.2.3 p. 484:


1] If a Normal topological space $X$ has a locally finite closed cover $(F\_f)\_{s\in S}$ and $dim F\_s\leq n\ s\in S$ then $dim X \leq n$.


from the subspace theorem ([ED]p.216):


2] For any subspace $M$ of a strongly-hereditarily-normal space (in particular a metric space) $X$ we have $dim M \leq dim X$


***In what follow assume that any open $\epsilon'$-ball ($\epsilon'\leq\epsilon$) B as $dim$-dimention $n$***. 


3] If $A\subset X$ contains a open $\epsilon'$-ball and is contained in a open $\epsilon'$-ball ($\epsilon',\epsilon''\leq \epsilon$) then $dim A=n$


PROOF: From [2].


4] There exist a locally finite covering of closed set $(F\_s)\_{s\in S}$ with $dim F\_s\leq n$.


PROOF: COnsider the covering by all open $\epsilon/2$-balls, and let $(B\_s)\_{s\in S}$ a locally finite refinement ($X$ is paracompact), let $F\_s:=Cl(B\_s)$ then $(F\_s)\_{s\in S}$ is a locally finite refinement of the open $\epsilon$-balls covering. Then as in [3] follow that $dim F\_s=n$.


Then from [1] and [4]: $dim(X)\leq n$ and from [2]: $n=dim B\_\epsilon(x) \leq dim X$


[E]: Engelking, General Topology
[ED]: Engelking, Dimention theory


