To cover a compact surface one needs at least two parametrizations.

Question

If $S$ is a compact surface then one needs at least two parametrizations to cover $S$

My try

I reason by contradiction. Suppose $X:U \to S$ is a parametrization that covers $S$, so that $X$ is an homeomorphism. The only argument I can think of uses that $U$ is open and $S$ is compact so that these sets can't be isomorphic because a compact set cannot be isomorphic to an open set.

In other to precise this claim I saw this question where they claim:

A non-empty open subset of $\mathbb{R}^n$ is never compact.

Perhaps I am being silly, but what is the formal justification of this fact?

score 2 · Accepted Answer · answered Apr 16 '18 at 22:24

2

In Hausdorff space, compact imply closed : How to prove that a compact set in a Hausdorff topological space is closed?

answered Apr 16 '18 at 22:24

LWW

756

So, in $\mathbb{R}^n$ a set cannot be open and closed at the same time? Because of connectedness right? – user1868607 Apr 17 '18 at 06:33
Javier / That's right. – LWW Apr 17 '18 at 06:51

To cover a compact surface one needs at least two parametrizations.

1 Answers1