# Existence of Borel cover of a metric space satisfying given properties

While reading a paper, I came across a claim without proof. Throughout the paper, $$X$$ is assumed to be a proper metric space, i.e., closed balls are compact (I don't know if this will be necessary for proving this claim).

Fix $$\varepsilon>0$$. There is a cover $$\{X_i\}$$ of $$X$$ such that $$(i)$$ $$X_i\cap X_j=\varnothing$$ if $$i\neq j$$, $$(ii)$$ $$\operatorname{diameter}(X_i)\leq\varepsilon$$ for all $$i$$, and $$(iii)$$ each $$X_i$$ has nonempty interior.

Here's my attempt at a proof:

Let $$\mathcal A$$ denote the set of all collections $$\{X_i\}$$ of Borel subsets of $$X$$ such that properties $$(i)$$, $$(ii)$$, and $$(iii)$$ above hold. Clearly $$\mathcal A$$ is nonempty. Ordering $$\mathcal A$$ by inclusion, the hypotheses of Zorn's lemma are satisfied, so there is a maximal element $$G=\{X_i\}$$ in $$\mathcal A$$.

Suppose $$\cup G\neq X$$, i.e., there is some $$x_0\in X$$ such that $$x\notin\cup G$$. Then either

(a) there is some $$\delta>0$$ such that $$B(x_0,\delta)\subset X\setminus(\cup G)$$, or

(b) for each $$n\in\mathbb N$$ there is some $$x_n\in\cup G$$ such that $$d(x_0,x_n)<\frac{1}{n}$$.

If (a) holds, then the maximality of $$G$$ is contradicted, so assume (b) holds.

And I seem to have hit a dead end. We could add $$x_0$$ to one of the $$X_i$$ and $$G$$ would still satisfy $$(i)$$, $$(ii)$$, and $$(iii)$$, but doing this for all elements in $$X\setminus(\cup G)$$ might make some set non-Borel.

Any help would be greatly appreciated.

• @mathrookie No advisor yet, but I'm trying to get into the noncommutative geometry group at Texas A&M, and thus have to read his papers. – Aweygan Mar 5 at 18:53
• ，Are you an undergraduate student from Texas A&M? – math112358 Mar 6 at 5:38

Here is one way to do this. Fix $$x_0 \in X$$. The set of annulus $$A_n=\{y \in X | n\leq d(x_0,y) \leq n+1\},$$ are compact (since $$X$$ is compact), so looking at the covering by open balls $$A_n \subset \cup_{x\in A_n} B(x,\epsilon),$$ we can extract a finite family $$(x_i)$$ such that $$A_n \subset \cup_i B(x_i,\epsilon).$$ Doing this for every $$n$$, since $$X=\cup_n A_n$$, we get a countable family (still denoted by $$x_i$$) of balls covering $$X$$, $$X=\cup_i B(x_i,\epsilon).$$ We now put $$Y_1=\overline{B}(x_1,\epsilon)$$ the closed ball of around $$x_1$$. $$Y_1$$ is of nonempty interior since $$x_1$$ is in its interior. We define a sequence of sets $$Y_n$$ recursively: assuming $$Y_1,...,Y_n$$ are defined in such a way that $$Y_1 \cup ... \cup Y_n$$ is a closed set, consider the set $$A=\overline{B}(x_{n+1},\epsilon)-(Y_1 \cup...\cup Y_n).$$ If it is empty or has non-empty interior, put $$Y_{n+1}=A$$. If $$A$$ is not empty but has empty interior, it means that there is a point $$y \in A$$ which is in $$\overline{B(x_{n+1},\epsilon)^c \cup Y_1 ...\cup Y_n}=\overline{B(x_{n+1},\epsilon)^c} \cup (Y_1 \cup...\cup Y_n).$$ But since $$y\notin (Y_1 \cup...\cup Y_n)$$, it must be a point such that $$d(x_{n+1},y)=\epsilon$$. In this case, we take $$Y_{n+1}= \overline{B}(x_{n+1},2\epsilon)-\cup Y_n,$$ and it has $$y$$ as interior point. And still, $$Y_1\cup ...\cup Y_{n+1}$$ is closed.
This way we have a sequence of sets such that $$\cup_{i=1}^n B(x_i,\epsilon) \subset \cup_{i=1}^n Y_i,$$ so $$\cup_{i\geq 0} Y_i=X$$, each $$Y_i$$ is either empty or of non-empty interior, and of diameter $$\leq 4\epsilon$$. All these sets are clearly Borel, and disjoint. Now remove the empty $$Y_n$$'s from the sequence, to get a sequence $$X_n$$ as desired.