By Definition, smooth manifolds are assumed to be Hausdorff and to satisfy the second countability axiom.
I have heard (but never seen written) that these assumptions imply paracompactness (and thus the existence of a Riemannian metric by the well-known construction using Partition of unity).
Does anybody know a reference or Proof for paracompactness?
Algebraic Topology – Why Are Manifolds Paracompact?
at.algebraic-topologydg.differential-geometrygt.geometric-topologysmooth-manifolds
Best Answer
Theorem: A countable atlas of charts for a Hausdorff $n$-manifold $M$ can be refined to a locally finite atlas. In fact, each chart only needs to be trimmed.
Proof: Let $U_1,U_2,\ldots$ be the charts. Each $U_i$, as a subset of $\mathbb{R}^n$, is the limit of a nested sequence of compact subsets $K_{i,1} \subseteq K_{i,2} \subseteq \ldots$. Since $M$ is Hausdorff, each $K_{i,j}$ is closed in $M$. So it suffices to delete $K_{1,i} \cup \cdots \cup K_{i-1,i}$ from $U_i$ to make a new chart $V_i$. Some of the $V_i$ might be empty, but this is no problem.