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?
That may not be true. Consider the graph of the function $x\mapsto \sin \frac{1}{x}$ over $(0, \infty)$. This is a complete submanifold of $\mathbb{R}^2$. If you take the product of this curve with $\mathbb{R}$ you get a complete surface in $\mathbb{R}^3$ which is not closed. Note that the surface is isometric to $\mathbb{R}^2$, very creased.