The Besicovitch Covering Property and Topological Dimension

Question

First of all, let me introduce some terminology. Here the notion of Topological Dimension is the one introduced by Lebesgue, it is also knwon as the Lebesegue Covering Dimension.

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Definition 1. A topological space $X$ is said to be finite dimensional if there is some $m\in\mathbb N$ such that for every open covering $\mathcal{A}$ of $X$, there is an open covering $\mathcal{B}$ of $X$ that refines $\mathcal{A}$ and has multiplicity $m+1$. The topological dimension of $X$ is defined to be the smallest value of $m$ for which this statement holds; we denote it by $ dim(X)$.

Definition 2. Let $(X,d)$ be a metric space. We say that $d$ has the Besicovitch Covering Property (BCP) if there exists an integer $N\ge1$ such that for every \textit{bounded} $A\subseteq X$ and every family $\mathcal{F}$ of closed metric balls of $X$ such that every point $a\in A$ is the centre of some balls belonging to $\mathcal{F}$, there exist a countable subfamily $\mathcal{G}$ of multiplicity $\le N$ covering $A$, namely,

\begin{equation} \chi_A\le\sum_{G\in\mathcal{G}}\chi_G\le N. \end{equation}

Here, if $X$ is a set and $S$ a subset of $X$, we denote by $\chi_S$ the characteristic function of $S$, that is to say the function such that $\chi_S(x)=1$ if $x\in S$, $\chi_S(x)=0$ if $x\notin S$.

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My Problem.

I have difficulties to prove the following fact.

Proposition. Every metric space $(X,d)$ satisfying the BCP has finite topological dimension.

The Proof is of the above proposition is obvious in the case $(X,d)$ is bounded but I don't know how to proocede when $(X,d)$ is un-bounded. Maybe I can use the fact that, if $x_0$ is a fixed point of $X$, then $X=\bigcup_{n\in\mathbb{N}} B(x_0,n)$? Or maybe I can use in someway the fact that every metric space is paracompact? I don't know. Can anyone help me please?

I take this opportunity to ask if anyone knows a good book about Topological Dimension. I've never studied this branch of the Topology and now I'd like to start.

Answer 1

I will outline a direct argument, but first let me remark that a space with BCP is clearly separable (otherwise the definition fails for $\frac{\epsilon}{2}$-balls centered around some uncountable $\epsilon$-separated set).

Furthermore, it's known (thanks to Moishe Kohan in the comments, see Theorem 7.2.1 in Engelking's "General Topology.") that topological dimension is preserved under countable unions in the case of separable metrizable spaces, which lets you reduce to the bounded case (with the caveat that in the bounded case you should still be careful to distinguish between open and closed balls - in fact its not obvious to me there's an easy way to circumvent this distinction other than my argument below).

I also recommend Hurewicz and Wallman's book (which is a little old but still good), which I'm almost certain has that result as well.

Also, I don't know offhand if that proof needs the axiom of choice - if it doesn't, it might be preferable to the argument I'm about to give in some sense (though note my remark at the end).

With that all said, you can prove this directly. Here's an outline of how it goes:

There's a subtle issue at play with open vs closed balls, with the BCP using closed balls and topological dimension using open sets. I would therefore argue as follows:

1. BCP clearly implies WBCP1 from this previous question on the Weak Besicovitch Covering Properties.
2. WBCP1 implies WBCP1(open), which is the identical property but for open balls. (To see this, note if $\mathcal F$ is a Besicovitch family of open balls with non-empty intersection containing $x$, then each member of $\mathcal F$ contains a closed ball with the same center intersecting $x$.)
3. In the remark at the end of this answer to that question, we observed that the method of proof gave an equivalence between different characterizations of WBCP1. That same remark applies to WBCP1(open).
4. Given an arbitrary open covering $\mathcal A$ of $X$, let $\mathcal F$ consist of the family of all open balls of radius $\frac{1}{n}$ for any $n\in\mathbb N$ that are fully contained in some member of $\mathcal A$. Then $\mathcal F$ is a refinement of $\mathcal A$, whose centers consist of all of $X$. You may therefore (using #3 in the aforementioned equivalence) select a subfamily $\mathcal G$, also covering $X$, with multiplicity $N$. Therefore the topological dimension is at most $N-1$.

Remark

Notice that we actually just proved a stronger result, namely that the Weak Besicovitch Covering Property (WBCP1) with constant $N$ is already enough to imply topological dimension at most $N-1$. Unlike the BCP, the WBCP1 property occurs in non-separable spaces, so our method of proof here still has some advantage.