Every point of a manifold M has a neighborhood homeomorphic to an open subset of $R^n$

276 Views Asked by At

"Every point of M has a neighborhood homeomorphic to an open subset of $R^n$."

I would like to understand this definition a bit better. With a homeomorphism, I understand it to be a continuous map with a continuous inverse. But why just continuous why not also differentiable? And why do we map to open subsets of $R^n$, why not map to closed sets?

1

There are 1 best solutions below

5
On
  1. Homeomorphic is enough to have a topological manifold. If, in addition, it is a diffeomorphism, then you've got a smooth manifold. Both are interesting objects themselves.

  2. By neighborhood it is actually meant an open neighborhood, so you want (small enough) open subsets of the manifold to look like $\mathbb{R}^n$, since the properties studied are usually local (continuity, differentiability and so on).