Metric Space is Paracompact/Proof 2

Proof
Let $M = \struct {A, d}$ be a metric space.

Let $\sequence {C_\alpha}$ be an open cover of $M$ indexed by ordinals.

Let $\map {B_r} x$ be the open $r$-ball in $M$ around $x$.

For $n \in \N$, let $D_{\alpha n}$ denote the union of all the open $r$-balls $\map {B_{2^{-n} } } x$ such that:


 * $(1): \quad$ $\alpha$ is the smallest ordinal such that $x \in C_\alpha$
 * $(2): \quad$ $x \notin D_{\beta j}$ for $j < n$
 * $(3): \quad$ $\map {B_{3 \times 2^{-n} } } x \subseteq C_\alpha$

It is to be shown that $\sequence {D_{\alpha n} }$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$.

By construction, $D_{\alpha n}$ is a refinement of $\sequence {C_\alpha}$.

Let $x \in A$.

Then there exists a smallest ordinal $\alpha$ such that $x \in C_\alpha$.

Then there exists $n$ sufficiently large such that $(3)$ holds.

Then by $(2)$:
 * $x \in D_{\beta j}$

for some $j \le n$.

Thus $\sequence {D_{\alpha n} }$ covers $M$.

Now assume $x \in A$.

Let $\alpha$ be the smallest ordinal such that $x \in D_{\alpha n}$ for some $n$.

Choose $j$ such that $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$.

Let $i \ge n$.

As $i > n$, every one of the open balls of radius $2^{-i}$ used to define $D_{\beta i}$ has its center $y$ outside $D_{\alpha n}$

And because $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$, we have that $\map d {x, y} \ge 2^{-j}$.

But $i \ge j + 1$ and $n + j \ge j + 1$.

So:
 * $\map {B_{2^{-n - j} } } x \cap \map {B_{2^{-j} } } y = \varnothing$

Thus if $i \ge n$, then $\map {B_{2^{-n - j} } } x$ intersects no $D_{\beta i}$.

Let $i < n$.

Let us choose:
 * $p \in D_{\beta i}$
 * $q \in D_{\gamma i}$
 * $\beta < \gamma$

There exist points $y$ and $z$ such that:
 * $p \in \map {B_{2^{-j} } } y \subseteq D_{\beta i}$
 * $q \in \map {B_{2^{-j} } } z \subseteq D_{\gamma i}$

By $(3)$:


 * $\map {B_{3 \times 2^{-n} } } y \subseteq C_\beta$

But by $(2)$:


 * $z \notin C_\beta$

So:
 * $\map d {y, z} \ge 3 \times 2^{-i}$

and:
 * $\map d {p, q} > 2^{-i} \ge 2^{-n - j + 1}$

Thus if $i < n$, then $\map {B_{2^{-n - j} } } x$ intersects $D_{\beta i}$ for at most one $\beta$.

Thus $D_{\alpha n}$ is locally finite.

Then $D_{\alpha n}$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$.

Hence by definition $M$ is paracompact.