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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.

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  • $\begingroup$ @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. $\endgroup$ – Aweygan Mar 5 at 18:53
  • $\begingroup$ ,Are you an undergraduate student from Texas A&M? $\endgroup$ – math112358 Mar 6 at 5:38
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I am unsure if Zorn is really useful for such a construction.

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.

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