Existence of a "special atlas"?

Question

Let $M$ be a smooth manifold and $\mathcal{A} =\{(\varphi_j, U_j) \}_{j\in J}$ an atlas of $M$.

I know that exists a refinement $\mathcal{V}$ of the cover $\{U_j\}_{j\in J}$, such that $\mathcal{V}$ is locally finite.

Now I would like to demonstrate the following statement:

Let $M$ be a manifold and $\mathcal{A}$ be an atlas of $M$, then exists an atlas $\mathcal{B}$ and a family of compact sets $\{K_i\}_{i\in I}$, such that are satisfied the following conditions:

• $\bigcup\limits_{i \in I} K_i = M$,

• $\forall$ $i$ $\in$ $I$, $\exists$ $(\varphi,U)$ $\in$ $\mathcal{B}$, such that $K_i$ $\in$ $U$

• If $\mathcal{B} = \{(\varphi_i,U_i)\}_{i\in I} $, then the cover $\{U_i\}_{i\in I}$ is locally finite.

• $\forall$ $(\varphi,U)$ $\in$ $\mathcal{B}$, there is $(\psi,V)$ $\in$ $\mathcal{A}$, such that $U \subset V$.

Can anyone help me to construct this atlas and this family of compact sets?

Answer 1

One convenient way to do this (which in fact shows that the index set $I$ may be taken to be countable) is to use Lemma 1.9 from Warner's excellent book Foundations of differentiable manifolds and Lie groups. This lemma implies that each open cover of a manifold possesses a countable locally finite refinement consisting of open sets with compact closures. (As usual, the definition of manifold includes the demands that $M$ be second countable and Hausdorff; otherwise your assertion is false).

This produces, given your initial atlas $\mathcal{A}$, the atlas $\mathcal{B}=(U_i,\phi_i)_{i \in \mathbf{Z}_{>0}}$ for which $\overline{U_i}$ is compact for each $i=1,2,\dots$ and $\mathcal{B}$ refines $\mathcal{A}$. Thus the third and fourth conditions you ask for hold.

It remains to observe that you can shrink each $U_i$ slightly, to obtain sets $V_i$ such that $K_i=\overline{V_i} \subseteq U_i$ with $\bigcup V_i=M$. The sets $K_i$ are compact since $U_i$ has compact closure. This is a standard argument; see e.g. the proof of Theorem 36.1 from Munkres' Topology (which works also for countably infinite covers).

We obtain your conditions, with the following improvements: the set $I$ may be taken to be countable, and in the first condition, the interiors of the sets $K_i$ already cover $M$.