TITLE: Question about complete metric on manifolds
QUESTION [7 upvotes]: I've recently been wondering about whether non-complete metrics on manifolds can be transformed into complete metrics on manifolds and whether all manifolds have complete metrics. After some googling I came across this link and the first comment says that any metric is actually conformal to a complete metric. I was wondering if anybody can show me a proof of this because I have had difficulty finding one. Thank you!

REPLY [6 votes]: Another way to argue that every second countable differentiable manifold $M$ admits a complete Riemannian metric is the following:
By Whitney, $M$ can be embedded into $\mathbb{R}^{2n+1}$ as a closed submanifold. The pullback metric on $M$ from $\mathbb{R}^{2n+1}$ then is complete since closed subsets of complete metric spaces are complete.