4
$\begingroup$

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 $\dim X\le n$?

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.