Please check the proof of the following theorem

Question

Let $X$ be a metrizable space.If $\mathcal A$ is an open covering of X,then there is an open covering $\varepsilon $ of $X$ refining $\mathcal A$ that is countably locally finite.

It is here we will use well-Ordering theorem.Pick a well ordering $<$ of the elements in $\mathcal A$.

Choose a metric $d$ for $X$.

Let $n$ be a positive integer,fixed for the moment.

Given an element $U$ of $\mathcal A$, let us define $S_n(U)$ as

$$S_n(U) = \{x:B(x,\frac{1}{n})\subset U\}$$

For each $U\in \mathcal A$, define $$T_n(U) = S_n(U)-\bigcup_{V<U} V$$

Now we will show that if $W$ and $V$ are distinct elements of $\mathcal{A}$, then $d(T_n(V),T_n(W)) \ge \frac{1}{n}$.

Assume WLOG that $V < W$, in the linear well-order we chose.

Proof: Let $x\in T_n(V)=S_n(V)-\cup_{{V'}<V} V'$ which in particular implies that $x \in S_n(V)$ and this means that $B(x,\frac{1}{n}) \subseteq V$.

Let $y\in T_n(W)= S_n(W)-\cup_{{V'}<W} V'$ which implies that $y \notin V$, as it is one of the sets we subtract to form $T_n(W)$ (and we know $y \in T_n(W)$ so that $y$ is not in $\frac{1}{n}$-neighbourhood of $x$, or it would have been in $V$ by the previous paragraph.

Hence, whenever $x\in T_n(V)$ and $y\in T_n(W)$, we have $d(x,y)\ge \frac{1}{n}$.

The sets $T_n(U)$ are not yet the ones we want, for we do not know that they are open sets. So, let us expand each of them slightly to obtain an open set $E_n(U)$. Specifically,let $E_n(U)$ be the $\frac{1}{3n}$-neighbourhood of $T_n(U)$. i.e. let $E_n(U)$ be the union of the open balls $B(x,\frac{1}{3n})$, for $x \in T_n(U)$. Mathematically, $$E_n(U) =\bigcup\{ B(x,\frac{1}{3n}): x\in T_n(U) \} $$.

($\implies T_n(U)\subset E_n(U)$)

Let $x\in E_n(V),y\in E_n(W),v\in T_n(V),w\in T_n(W)$,then $d(v,w)\ge \frac{1}{n}.$So,$\frac{1}{n}\le d(v,w)\le d(v,x)+d(x,y)+d(y,w)\implies d(x,y)\ge \frac{1}{n}-d(v,x)-d(y,w)\ge \frac{1}{n}-\frac{1}{3n}-\frac{1}{3n}=\frac{1}{3n}$.

$(d(v,x) \le \frac{1}{3n})$

So, $ d(E_n(V),E_n(W)) \ge \frac{1}{3n}$.

let us define: $$\varepsilon_n=\{ E_n(U): U \in \mathcal A \}$$

Claim:$\varepsilon_n$ is locally finite collection of open sets that refines $\mathcal A$.

• $\varepsilon_n$ refines $\mathcal{A}$ comes from the fact that $E_n(V)\subset V$ for each $V\in \mathcal{A}$.
• $\varepsilon_n$ is locally finite comes from the fact that for any $x \in X$, the $\frac{1}{6n}$-neighbourhood of $x$ can intersect at most one element of $\varepsilon_n$.(1.How to prove this fact?)
• $\varepsilon_n$ will not cover $X$(2.How to prove this fact?).

Claim:$\varepsilon=\cup_{n\in \mathbb Z_{+}} \varepsilon_n$ covers $X$.

Let $x \in X$. The collection $\mathcal A$ with which we began covers $X$.Let us choose $U$ to be the first element of $\mathcal{A}$( in the well ordering,$<$)that contains $x$. Since $U$ is open ,we can choose $n$ so that $B(x,\frac{1}{n})\subset U$. So, $x\in S_n(U)$. Now,because $U$ is the first element of $\mathcal{A}$ that contains $x$, the point $x \in T_n(U)$ implies that $x$ belongs to the element of $E_n(U)$ of $\varepsilon_n \subseteq \varepsilon$. Hence,$\varepsilon$ covers $X$ .$\blacksquare$

Answer 1

• You should start out by explicitly chosing a well-order $<$ on the open cover $\mathcal{A}$. This hints (which is true) that the theorem essentially uses the axiom of choice (which I personally have no problem with, but some people might).

• The remark that the $\varepsilon_n$ need not cover $X$ is just pedagogical. What only matters is that the union of all these does cover $X$, which is your final point.

• I find the proof that $d(E_n(V),E_n(W)) \ge \frac{1}{3n}$ unclearly written. Try and improve that (how do you introduce your variables e.g.)

• To see your $1$, use what was shown above it: namely that $d(E_n(V), E_n(V)) \ge \frac{1}{3n}$ whenever $U \neq V$. If a ball $B(x, \frac{1}{6n})$ would intersect two members of $\varepsilon_n$, it would intersect say $E_n(V)$ (in $p$ say) and $E_n(W)$ (say in $q$) and then computing the distance between these intersection points via $x$, shows that $$d(p,q) \le d(p,x) + d(q,x) < \frac{1}{6n} + \frac{1}{6n} = \frac{1}{3n}$$ and thus $d(E_n(V), E_n(V)) < \frac{1}{3n}$ contradicting what we already knew. So every $x \in X$ has a neighbourhood $B(x, \frac{1}{6n})$ that intersects at most one member of $\varepsilon_n$ (so $\varepsilon_n$ is even discrete, not just locally finite, and we have a countably discrete cover $\varepsilon$).

• You could explain $E_n(V) \subseteq V$ better as well. Why does that hold? It's true but a proof should explain it.

• You could end with "and this concludes the proof", or quod erat demonstrandum or ὅπερ ἔδει δεῖξαι etc. instead of so abruptly.

This proof is essentially the same one as given in Munkres (2nd edition), p246-247, so no credits for originality.