- #1

Gene Naden

- 319

- 64

Here is my attempt at a solution. At a key step it makes an assertion without proof.

Consider a point p in M-R where M is the surface and R is the region in M. To show that p has a neighborhood outside of R, we need a lower bound on the distance of p from R.

For every ##\epsilon>0## the set of all neighborhoods of points in R covers R, so the neighborhoods of a finite set S of points in R covers R. Let q be the point in S that is closest to p. The point in R that is closest to p might be in the ##\epsilon## neighborhood of q. If so, the distance of p from R is greater than or equal to ##d(p,q)-\epsilon## where d is distance. If we choose ##\epsilon<d(p,q)## then the distance of p from R is greater than zero, and there is a neighborhood of p in ##M-R## and so M-R is open and R is closed.