# Every complete regular surface is closed?

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?

• One possible result is: if $M$ complete is an open submanifold of $X$, then $M$ must be a union of connected components of $X$. In particular, an open complete submanifold of $\mathbb{R}^n$ is $\mathbb{R}^n$. That may not be enough for what you want to prove. Oct 9 '17 at 10:26
• I made a mistake in my original post. Even if we were able to show that $(M,\tilde{d})$ is complete, that would not necessarily mean that $(M,d_{|M})$ is complete. That $\tilde{d}$ dominates $d_{|M}$ means that every Cauncy squence in $(M,\tilde{d})$ is a Cauchy sequence in $(M,d_{|M})$. This means that it is the completess of $(M,d_{|M})$ that implies the completeness of $(M,\tilde{d})$; not the other way around. Oct 12 '17 at 21:05
• Yes, that was the problem I faced, did try to prove it and couldn't.... There were these sequences going nowhere in $M$ and still Cauchy in $\mathbb{R}^n$. A good question nevertheless. Oct 12 '17 at 21:19
• @orangeskid Is the reverse implication true? I.e. is every closed regular surface geodesically complete? From point-set topology, we know that $M$ being a closed subset of $\mathbb{R}^3$ means that $(M,d_{|M}$ is a complete metric space, which, as we have seen above, implies that $(M,\tilde{d})$ is also a complete metric space. Can (a variation of) Hopf-Rinow now be applied to conclude that $M$ is geodesically complete in the sense of the definition in my original post? Oct 12 '17 at 23:05
• Yes., geodesically complete is equivalent to $\tilde d$ complete. That is part of Hopf-Rinow if I am not mistaken. Of course, it has to be a closed submanifold of a complete manifold. I don't have a good source for Hopf-Rinow. Maybe Bishop and Crittenden? Oct 12 '17 at 23:28

## 1 Answer

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.

• Interesting, so although a the distance from a fixed point to a limit point is finite in $\Bbb R^3$, the intrinsic distance is going to infinity as you approach it. I was going to suggest writing $p_n = \exp_p (t_nv_n)$, where the $v_n$ are unit vectors in $T_pM$. You certainly can assume $v_n\to v$, but the $t_n\to\infty$. Oops. (I think the same thing will happen even with $x\sin(1/x)$, as it is also of infinite arclength.) Oct 8 '17 at 22:01
• @Ted Shifrin; Yes indeed. Would be nice to plot some geodesics. Oct 8 '17 at 22:23
• Well, this is just a cylinder over your plane curve. So we know what the geodesics are on any such creature. Oct 8 '17 at 22:25
• @Oskar Henriksson: The induced metric is in fact a product metric. As for the factors, one is the standard $\mathbb{R}$, the other one is the long graph ( so again isometric to $\mathbb{R}$). A cylinder in fact, over this long curve. So it's isometric to $\mathbb{R}^2$. If you consider the less wild curve ( still of infinite length) suggested by Ted Shifrin ( graph of $u \sin \frac{1}{u}$, you can take both of the pieces. They are then two $\mathbb{R}^2$, connected at a singular edge. Funny... Oct 12 '17 at 21:36
• @Oskar Henriksson: then you have a topological manifold, with a singular edge, and the smooth pieces are complete with the induced metric. Perhaps even nicer. And you can never reach the edge if the speed is bounded... Oct 12 '17 at 21:38