Equivalent condition for refinement of open coverings of a regular space

Question

As per Topology Munkres (Second Edition) there is a Lemma (41.3) (I am writing required part only : Let $X$ be a regular space. If every open covering of $X$ has a refinement that is an open covering of $X$ and countably locally finite then every open covering of $X$ has a refinement which is a covering for $X$ that is locally finite.

The proof starts with an open covering $\mathcal{A}$ of $X$ and its countably locally finite open refinement $\mathcal{B}= \bigcup \mathcal{B}_{n}$ which cover $X$. $V_{i}= \displaystyle \bigcup_{U \in \mathcal{B}_{i}} U$, $S_{n}(U)= U - \displaystyle \bigcup_{i < n} V_{i}$. Collection of such $S_{n}(U)$ is denoted by $\mathcal{C}_{n}$ and $\mathcal{C}= \bigcup\mathcal{C}_{n}$.

For locally finiteness of $\mathcal{C}$ : Let $x \in X$ and $N$ be the smallest integer such that $x \in U \in \mathcal{B}_{N}$. Hence $x \in S_{N}(U)$. For each $n=1,2..N$ we get neighbourhoods $W_{n}$ of $x$ which intersects finitely many elements of $\mathcal{B}_{n}$ and these $W_{n}$ can intersect only finitely many elements of $\mathcal{C}_{n}$. Also $U$ does not intersect no element of $\mathcal{C}_{n}$ for $n>N$. Thus the neighbourhood $W_{1} \cap \cdots \cap W_{n} \cap U$ is a neighbourhood of $x$ which intersects finitely many elements of $\mathcal{C}$.

My doubt is this elements of $\mathcal{C}$ is actually elements of $\mathcal{C}_{n}$. First neighbourhood $W_{1}$ intersects finitely many elements of $\mathcal{C}_{n}$. Is it sufficient for locally finiteness of $\mathcal{C}$. If any of the $W_{n}$ does not intersect corresponding $S_{n}(U)$'s then how can $W_{1} \cap \cdots \cap W_{n} \cap U$ intersect elements of $\mathcal{C}$ ?

Answer 1

Here is the same proof in slightly expanded form (filling all details).

Note that the proof of local finiteness of $\mathcal{C}$ goes hand in hand with the proof it is a cover. By $S_n(U) \subseteq U$ for all $U$ it is clearly a refinement of $\mathcal{B}$ and so of $\mathcal{A}$ as well.

So let $x \in X$.
There is clearly a first $n$ such that $x$ is covered by some $\mathcal{B_n}$, as $\mathcal{B} = \cup_n \mathcal{B}_n$ is a cover. So the set $\{ n \in \mathbb{N}: x \in \cup \mathcal{B}_n= V_n\}$ is a non-empty subset of $\mathbb{N}$ so has a minimum $N$. Also, $x \in U$ for some $U \in \mathcal{B}_N$. By the minimality:

$$\forall i < N: x \notin V_i$$

so that $x \in S_N(U) = U - \cup_{i < N} V_i$.

Note that $U$ is an open neighbourhood of $x$ that intersects no $C \in \mathcal{C}_n$ for $n > N$. This is a first step towards showing that $\mathcal{C}$ is locally finite at $x$: for that we need a neighbourhood of $x$ that intersects only finitely many elements of $\mathcal{C} = \cup_n \mathcal{C}_n$. At least we know that $U$ already intersects no elements from the tail $\cup_{n > N} \mathcal{C}_n$.

To see this last claim: suppose $S_n(V) \in \mathcal{C}_n$ for some $n > N$ and some $V \in \mathcal{B}_n$. If $p$ were in $S_n(V) \cap U$ then $p \in U \subseteq V_N$, so $p \notin S_n(V)$, because $n > N$ means that we threw out all $V_i$ for $i < n$ and $V_N$ is one of those.

So the idea is now to find a smaller neighbourhood of $x$ inside $U$ that takes care of all $\mathcal{C}_i$ for $n \le N$ as well: there are only finitely many conditions, so for each of those $n \le N$, we find an open neighbourhood $W_n$ of $x$ such that

$$F_n:= \{B \in \mathcal{B}_n: W_n \cap B \neq \emptyset \} $$

is finite, which can be done as each collection $\mathcal{B}_n$ is locally finite by assumption. Now define $O_x = U \cap W_1 \cap \ldots \cap W_n$, which is an open neighbourhood of $x$ (as we have a finite intersection of open sets).
Suppose $O_x \cap S_n(U) \neq \emptyset$ for some $U \in \mathcal{B}_n$ (this is the general form of an element of $\mathcal{C}$). We know that $U \cap S_n(U) = \emptyset$ when $n > N$ by the first part. So $n \le N$ must hold. But $S_n(U) \subseteq U$ so $O_x \cap U \neq \emptyset$, for this $U \in \mathcal{B}_n$, and in particular (as $O_x \subseteq W_n$ for this $n$ by construction)

$$W_n \cap U \neq \emptyset$$

which means that $U \in F_n$, one of the finitely many elements of $\mathcal{B}_n$ that intersects $W_n$. This means that $O_x$ can only intersect the $S_n(U)$ for $U \in \cup_{n \le N} F_n$, a finite set (a finite union of finite sets). So $O_x$ shows that $\cup_n \mathcal{C}_n =\mathcal{C}$ is locally finite at $x$. As $x$ was arbitrary we are done.