Example of complete but not closed Riemannian submanifold

Question

I am working on an exercise where I am asked to construct an example of a complete Riemannian metric $(M,g)$ and a connected embedded Riemannian submanifold $P \subseteq M$ that is complete, but not closed. And I am not certain that such an example exists.

Suppose $x \in M \setminus P$ is a limit point of $P$. Then there is a sequence $\left\{x_n\right\}_{n \geq 1} \subset P$ that converges to $x$ in the metric $d_g$ induced by the Riemannian metric on $M$. In particular, $\left\{x_n\right\}$ is a Cauchy sequence in $M$. So in order for this to be true, we would need $\left\{x_n\right\}$ to fail to be a Cauchy sequence in $P$; otherwise by completeness of $P$, $x_n\to x \in P$ in the induced topology on $P$, contradicting our hypothesis.

What I need: I want an example of a complete Riemannian manifold $(M,g)$ admitting a connected, complete, embedded submanifold $P \subset M$ where Cauchy sequences in $M$ are not necessarily Cauchy in $P$. But I'm not sure how to construct such a thing. Any tips?

Answer 1

Your goal is to have points which are arbitrarily close in $M$ but far apart in $P$. In other words, you want to have points that get close together in $M$ but such that to get between them in $P$, you have to travel a large distance. Intuitively, this is easy: for instance, you can draw a curve that keeps revisiting a location infinitely often (getting closer and closer each time) while moving a fixed distance away each time.

A specific example of such a curve is hidden below.

Let $M=\mathbb{R}^2$ with its usual metric, and let $P=\{(x,\sin(1/x)):x>0\}$. As $x$ approaches $0$, $\sin(1/x)$ oscillates between $-1$ and $1$ infinitely often, so the curve accumulates at $(0,y)$ for all $y\in[-1,1]$ and $P$ is not closed. But $P$ is complete in its induced metric, since the arc length required to approach $x=0$ is infinite.