Every metric space is paracompact (an elegant proof)

Question

I'm studying a different proof to show that each metric space is paracompact in the book: Singh, Tej Bahadur-Introduction to Topology. It is a very elegant construction unlike the inductive method that seems more cumbersome to me. However there are three things that I could not understand:

(1) Why can some $E_{n, \alpha}$ be empty?

(2) How can I intuitively or geometrically see the sets $F_{n,\alpha}$ and $V_{n,\alpha}$?

(3) Why is it obtained that $F_{n,\alpha} \subset X – U_\beta $ or $F_{n,\beta} \subset X – U_\alpha$ in the last part of the proof?

I'm very interested in knowing a good argument for these questions since I have not been able to figure it out on my own after many attempts and I'm eager to know the answers. Any help is appreciated.

Here are some definitions:

Definition.
Let $X$ be a topological space. A collection of sets $\{U_{\alpha}\} \subset X$ (not necessarily open or closed) is said to be locally finite if to each ${x \in X}$, there is a neighborhood ${U}$ of ${x}$ that intersects only finitely many of the ${U_{\alpha}}$

Definition. Let ${\left\{U_{\alpha}\right\} }$ be a cover of a space ${X}$. Then a cover ${\left\{V_{\beta}\right\}}$ is called a refinement if each ${V_{\beta}}$ sits inside some ${U_{\alpha}}$

Definition. A Hausdorff space is paracompact if every open covering has a locally finite refinement.

Answer 1

1. For each $\alpha$, removing the points of $U_{\alpha}$ at distance $< 1/n$ from the complement of $U_{\alpha}$ leaves $E_{n,\alpha}$. Loosely, trim away an open strip of width $1/n$ around the boundary of $U_{\alpha}$. (It may help to note that the family $\{E_{n,\alpha}\}_{n=1}^{\infty}$ is an exhaustion of $U_{\alpha}$, i.e., the sets are nested and their union is $U_{\alpha}$.) The set $E_{n,\alpha}$ is empty if every point of $U_{\alpha}$ is at distance $< 1/n$ from the complement of $U_{\alpha}$. Qualitatively, the set $U_{\alpha}$ is "thin" in the sense of containing no closed ball of radius $1/n$.
2. Well-ordering of $A$ means the covering sets can be ordered so that there is a "first" set, and each set $U_{\alpha}$ has a successor (unless there is a "last" set). Now, think of $E_{n,\alpha}$ as a cookie, and pass $E_{n,\alpha}$ along the list of covering sets preceding $U_{\alpha}$, with each set $U_{\alpha'}$ ($\alpha' \prec \alpha$) taking a bite (subtracting $U_{\alpha'}$, i.e., intersecting with $X - U_{\alpha'}$). The set $F_{n,\alpha}$ is what remains; it contains points of $U_{\alpha}$ that are at distance $\geq 1/n$ from the complement of $U_{\alpha}$ and do not lie in $U_{\alpha'}$ for every $\alpha' \prec \alpha$. Place an open ball of radius $1/3n$ about each point this partially-eaten cookie $F_{n,\alpha}$; the union of these balls is $V_{n,\alpha}$.
3. If $\alpha$ and $\beta$ are two distinct indices in $A$, one comes before the other, say $\alpha \prec \beta$ without loss of generality. By construction, the open set $U_{\alpha}$ took a bite out of $F_{n,\beta}$, so $U_{\alpha} \cap F_{n,\beta} = \varnothing$, i.e., $F_{n,\beta} \subset X - U_{\alpha}$.