Visualization of the proof in Lee's Smooth Manifolds book that every open cover has a regular refinement

Question

Here is the theorem Lee is proving.

(It states that any open cover on a smooth manifold has a regular refinement (a refinement which is countable, locally finite, and satisfies additional ad-hoc properties). The proof is based on a previously proven lemma which states that any smooth manifold has a countable, locally finite open cover with paracompact sets.)

I can understand everything except for the last paragraph. In particular, I am unable to visualize the situation in the last paragraph. Can anyone help me visualize the situation? Thanks!

Answer 1

Maybe it's easier to see if we change the given precompact open sets and $\textit{then}$ define the charts. For this purpose, we can employ a trick, which I think I learned from Munkres in his book, Analysis on Manifolds. Of course, if $M$ is compact, there is nothing to do. Otherwise, set $K_{-1}=K_0=\emptyset$ and $K_1=\overline V_1$ and if $\{K_i\}^n_{i=1}$ have been chosen, let $m$ be the first integer greater than $n$ such that $K_n\subseteq \bigcup^m_{i=1}V_i$ and define $K_{n+1}=\bigcup^m_{j=1}\overline V_j.$

In what follows, I recalled the proof (I hope correctly!) just by drawing a picture of the expanding $K_{i-1},K_i,K_{i+1},K_{i+2}$ and their interiors, and overlaying the charts.

For each $A_{\alpha}\in \Xi,\ $ the given open cover, construct the open set $(\text{int}K_{i+2}\setminus K_{i-1})\cap A_{\alpha}$ and for each $p\in (\text{int}K_{i+2}\setminus K_{i-1})\cap A_{\alpha},\ \textit{now}$ take charts $(U_{p,\alpha},\psi_{p,\alpha})$ such that $U_{p,\alpha}\subseteq (\text{int}K_{i+2}\setminus K_{i-1})\cap A_{\alpha},\ \psi_{p,\alpha}(p)=0$ and $\psi_{p,\alpha}(U_{p,\alpha})\subseteq B_3(0).$

If we let $W_{p,\alpha}=\psi^{-1}_{p,\alpha}(B_1(0))$, then $W_{p,\alpha}\subseteq(\text{int}K_{i+2}\setminus K_{i-1})\cap A_{\alpha}$. Clearly, $\{W_{p,\alpha}:p\in M,\alpha\in J\}$ refines $\Xi$. And, by construction, $K_{i+1}\subseteq \text{int}K_{i+2}$ so from the collection $\{W_{p,\alpha}:p\in M,\alpha\in J\}$ we get a finite covering of the compact set $K_{i+1}\setminus \text{int}K_i.$ As this is true for each integer $i$, we obtain a countable sequence of sets $\{(W'_i,\psi'_i)\}^{\infty}_{i=1}$ and these are the sets that work. Indeed, if $p\in M$ then there is an integer $i$ such that $p\in \text{int}K_i$, which by construction, can intersect only finitely many $W'_i$.

Answer 2

I had problems with the last paragraph too when following this proof. Here is how I understand it (I use $\left\lbrace V_j \right\rbrace$ to mean the entire cover provided by the lemma he proves before, not the set with the single element):

1. Every $W_{i,j}$ is also a $W_p$ for some $p \in M$; since $\left\lbrace V_j \right\rbrace$ is a cover, there exists a $V_{s(i,j)}$ such that $p \in V_{s(i,j)}$ and hence, by the third bullet point, $W_{i,j} \subset V_{s(i,j)}$.
2. Fix now a $W_{k,i}$. Claim: $W_{k,i}$ intersects only with finitely many sets in the cover $\left\lbrace V_j \right\rbrace$. Explanation: because $\left\lbrace V_j \right\rbrace$ is locally finite, $V_{s(k,i)}$ intersects with finitely many elements in $\left\lbrace V_j \right\rbrace$ (see Exercise 2.9, in which this equivalent characterisation of local finitude is given). Since $W_{k,i} \subset V_{s(k,i)}$, the claim follows.
3. Claim 2: for each $V_s$ in the cover $\left\lbrace V_j \right\rbrace$, there are at most finitely many sets $W_{l, j}$ which have non-empty intersection with $V_s$. Justification: if, for a given $s$, an infinite number of $W_{l,j}$'s would intersect $V_s$, then so would the elements of $\left\lbrace V_j \right\rbrace$ associated to them i.e. there would be an infinity of $V_{s(l,j)}$'s intersecting $V_s$, contradiction.
4. So by 2. every $W_{k,i}$ intersects finitely many sets in $\left\lbrace V_j \right\rbrace$ and by 3. it follows that $W_{k,i}$ intersects at most a finite number of other $W_{l,j}$'s.

There doesn't seem (to me) to be a need for covering $\overline{V_k}$ with finitely many $V_j$'s i.e. a need for the compactness of $V_j$, but I wouldn't worry about that, because the compactness of each $\overline{V_k}$ is used crucially in the paragraph above.

I hope this helped.