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 $\dim X\le n$?


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

  • $\begingroup$ Thank you very much. It is strange that such basic statement needs to be proved. I thought it should be a standard theorem. $\endgroup$ – ε-δ Nov 7 '10 at 4:41

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