# Does every open cover of a paracompact Hausdorff manifold admit countable refinement?

I've begun studying manifolds, and an exercise asked to prove the existence of subordinate partitions of unity. One of the steps is this question:

Let $M$ be a paracompact, Hausdorff, smooth manifold. Show that every open cover of $M$ admits a countable, locally finite refinement by open sets.

Of course, the "locally finite" part comes directly from the definition of paracompactness. However, I'm stuck in the "countable" part. It's never implied that the manifold is second-countable, which almost all my search results use as hypothesis, so I either don't need it or I have to prove.

If this isn't related to manifolds, but it's a direct result of Hausdorfness, paracompactness or both, I'd like a proof as well.

• Are manifolds defined as connected? If not, the assertion is wrong, consider a manifold with uncountably many connected components, and the open cover given by the components. And a (Hausdorff) manifold is paracompact if and only if each connected component is second countable. – Daniel Fischer Aug 23 '17 at 20:58
• I don't think so. Would smoothness imply connectedness, or is there a counterexample? – AspiringMathematician Aug 23 '17 at 21:39
• Smoothness wouldn't imply connectedness, $\mathbb{R}\times \{-1,1\}$ is in a natural way a smooth manifold with two connected components. For the assertion to be true, it is necessary that the definition of a manifold says (directly or indirectly) that a manifold has at most countably many connected components. The usual ways to do that are 1. to define a manifold as second countable; but then every Hausdorff manifold is paracompact, so the exercise wouldn't make sense; or 2. to define a manifold as connected. Either there is something in the used definition of a manifold that guarantees only – Daniel Fischer Aug 24 '17 at 10:32
• countably many components, or the author overlooked the necessity of that condition when posing the exercise. Whatever the case, do the exercise under the assumption that $M$ is connected. The argument then applies to each component separately, and thus straightforwardly yields the case of countably many components. To deal with the connected case, this question provides an idea how to go about it. – Daniel Fischer Aug 24 '17 at 10:32
• Thanks for the clarification! I'll have a look if it was assumed elsewhere. – AspiringMathematician Aug 24 '17 at 16:39