Show that a connected 1d manifold $M$ is homeomorphic to $\mathbb{R}$ or $S^1$ [duplicate]

I know there are a couple of posts where this question has already been asked, but I did not see any proof from the following approach:

I have basically shown already that if I have two open sets that cover $M$, being each of them homeomorphic to $\mathbb{R}$ itself, then $M\cong \mathbb{R}$ or to the circle.

I am following the steps of this paper. Here the argument is divided in the case when $M$ is compact and when it is not. Also, I assume that the manifold has no boundary (otherwise the argument would fail anyways).

So my questions is: How does one prove the case when $M$ is compact, that it is homeomophic to a circle having proved what I mentioned? It seems that what I need is just some induction to conclude, but I am not sure how to do it. And also, could anyone give me some hints in how to finish off the proof when $M$ is not compact? Ideally, but of course not neccessarily, using the fact that I have already shown? Thanks a lot.

Solution 1:

Proposition 3.1 allows you to reduce a finite cover by $O$-sets to two, using induction on the number of finite $O$-sets in a cover of $X$.

Suppose (induction hypothesis) that a space that can be covered by $n$ $O$-sets (where $n \ge 2$) can in fact be covered also by two $O$-sets. Then if $X= \bigcup_{i=1}^{n+1} U_i$ is cover by $n+1$ $O$-sets, replace $U_n, U_{n+1}$ by their union $U_n \cup U_{n+1}$ which is an $O$-set by 3.1. We can apply the induction hypothesis (we now have $n$ sets) and we're done. The base case $n=2$ is already trivial.

So compactness plus boundary-lessness gives a finite cover by $O$-sets (we cannot have one set in that cover) so the fact we just showed by induction tells us that $X=U_1 \cup U_2$ with $U_1,U_2 \simeq \Bbb R$ and we're done.