How to construct point finite covering in collectionwise normal spaces?

How do I prove there always exists a subset A of a topological basis B of a manifold M, such that A is locally finite open covering of M?

• This problem is related to the book Lectures on Differential Geometry by S. S. Chern, etc.. In Chapter 3, there is a theorem. Theorem 3. It states that suppose ΣΣ\Sigma is a topological basis of the manifold M. There is a subset Σ0Σ0\Sigma_0 of ΣΣ\Sigma such that Σ0Σ0\Sigma_0 is a locally finite open covering of M. Let me first show my reasoning. By definition, if an open covering (call it Σ0Σ0\Sigma_0) of M satisfies the following condition that every compact subset of M intersects only finitely many elements of Σ0Σ0\Sigma_0, Σ0Σ0\Sigma_0 is the locally finite open covering of M. Therefore, we can prove Theorem 3 in the following way: ΣΣ\Sigma is an open covering of M, so it is also an open covering of any compact subset (call it C) of M. Since C is compact, there is a finite open covering ΣCΣC\Sigma_C, so C intersects finitely many elements of ΣΣ\Sigma. Therefore, ΣΣ\Sigma itself is a locally finite open covering of M. So this theorem looks trivial. But why do we need this theorem? In particular, Chern etc. gave a long proof. What is their point? Thanks!

• Answer:

  From "CCC has a finite open subcovering ΣCΣC\Sigma_C" it doesn't follow that "CCC intersects only finitely many elements of ΣΣ\Sigma". It could certainly intersect infinitely many of them. Just because you can pick a finite subcovering doesn't mean that any other set doesn't intersect your compact set CCC. You should note that your proof in the third paragraph shows something patently absurd: starting with an arbitrary open covering ΣΣ\Sigma, you show that ΣΣ\Sigma itself is locally finite. In other words, you apparently proved that every covering is locally finite. This is obviously untrue, neither for manifolds nor in general, as you should readily be able to produce counterexamples.