Problem. I'm working on an exercise from a book in basic Gaussian geometry, and as a part of my solution, I would like to make the following claim:
Proposition. If $M\subseteq\mathbb{R}^3$ is a complete regular surface, $M$ is a closed subset of $\mathbb{R}^3$.
given the following definition of completeness:
Definition. A regular surface $M\subseteq\mathbb{R}^3$ is said to be complete if for every point $p\in M$ and every tangent vector $Z\in T_pM$, there exists a geodesic $\gamma\colon\mathbb{R}\to M$ defined on all of $\mathbb{R}$ such that $\gamma (0)=p$ and $\gamma'(0)=Z$.
I have tried mainly two ways of proving it:
Idea 1. What seems most natural to me is to do this by contradiction, and to use the limit point characterization of closed sets in metric spaces.
I thus let $M\subseteq \mathbb{R}^3$ be a complete regular surface, and suppose that $M$ is not closed. This means there exists a convergent sequence $(p_n)_{n=1}^\infty$ in $\mathbb{R}^3$ with limit $q$, such that $p_n\in M$ for all $n\in\mathbb{Z}^+$ and $q\in\mathbb{R}^3\setminus M$.
The way I picture this intuitively, is that the surface $M$ is punctured at $q$ or have something like an "edge " at $q$. Since there are $p_n$'s arbitrarily close to $q$, it feels like we should be able to find a $p_n$ and a $Z\in T_{p_n}M$, such that the geodesic $\gamma\colon \mathbb{R}\to M$, with $\gamma (0)=p_n$ and $\gamma'(0)=Z$, would have to pass through $q$, thus giving us the desired contradiction. I don't see any obvious ways to turn this into a formal argument though, and my gut feeling can very well be wrong.
Idea 2. I have also looked a bit at the Hopf-Rinow theorem from Riemannian geometry, which seems to say that a Riemannian manifold $(M, g)$ being geodesically complete implies that $(M,\tilde{d})$ is complete as a metric space, where $\tilde{d}$ is the intrinsic metric.
If something similar holds for regular surfaces in $\mathbb{R}^3$, I think we would be done (no!), because $\tilde{d}$ dominates the standard metric $d$ in $\mathbb{R}^3$ restricted to $M$. This means that if $\rlap{\rule[0.5ex]{2.5em}{0.2ex}}(M,\tilde{d})$ is complete, so is $\rlap{\rule[0.5ex]{3.5em}{0.2ex}}(M,d_{|M})$ (this is wrong, see my comment below!). Since $(\mathbb{R}^3,d)$ is complete, $(M,d_{|M})$ being complete implies that $M$ is closed by elementary point-set topology.
However, my spontanious idea of how do show that geodesically complete implies complete as a metric space would be very similar to Idea 1 (I would seek a contradiction by supposing that we have a Cauchy sequence in $M$ that doesn't converge), so this doesn't seem to take me any further either.
Question. Am I at all on the right track here, or would you recommend some other approach? Is the proposition I'm trying to prove even correct in the first place?