What are some examples of incomplete nonextendible manifolds? A connected Riemannian manifold $(M,g)$ is said to be extendible if it is isometric
to a proper open subset of a larger connected Riemannian manifold (this is the definition given in Lee's Introduction to Riemannian manifolds). I was able to show that a complete Riemannian manifold is nonextendible. I would like to find an example of incomplete Riemannian manifold which is not extendible. According to Wikipedia such manifolds exist, but no reference is given.

What are some (relatively simple) examples of connected incomplete nonextendible Riemannian manifolds ?

I tried looking at the case of the real numbers $\mathbf{R}$, without great success. More specifically I tried to find conditions on a metric on $\mathbf{R}$ to be extendible. Suppose that there is a metric $g$ on $\mathbf{R}$ and an isometric embedding $\varphi:(\mathbf{R},g)\hookrightarrow (M,g_0)$ with $M$ a connected 1-manifold and with $\varphi(\mathbf{R})\subsetneq M$. For example we can look at the case $M=\mathbf{R}$ and $\varphi(\mathbf{R})=(-\infty,0)$ with $\varphi$ increasing. Writte $g_0 = f(x)dx^2,~f>0$. Then, $\varphi$ bieng an isometry, we get
$$g = \underbrace{f\circ\varphi(x)\varphi^\prime(x)^2}_{h(x):=}dx^2.$$
Now $\varphi$ is strictly increasing and $\lim_{x\to+\infty} \varphi(x)=0$ so there is a sequence $(x_n)_n\in\mathbf R^\mathbf N$ converging to $+\infty$ such that
$$\lim_{n\to\infty}\varphi^\prime(x_n)=0.$$
Since $f$ in bounded near $0$, we get
$$\lim_{n\to\infty}h(x_n)=0.$$
Therefore if we want to construct nonextendible metric $g=h(x)dx^2$ on $\mathbf R$ "by hand", we might want to assume $h$ is bounded bellow by a positive constant $\varepsilon>0$ (so that there is no sequence $(x_n)_n\in\mathbf R^\mathbf N$ with $\lim_{n\to\infty} \vert x_n\vert = +\infty$ and $\lim_{n\to\infty}h(x_n)=0$). However I am pretty sure that these metrics are all complete (even if I don't have an actual proof of this fact), so I don't think I am gonna find an example in this way.
In dimension two, the only incomplete manifolds I can think of are punctured surfaces, but these are all extendible.
Any kind of help will be greatly appreciated.
 A: Although this has been answered a while ago, I cannot resist describing one of my favorite examples, which is related to the comment of @TedShifrin.
Start with the ordinary Euclidean metric on $\mathbb R^2 - \{(0,0)\}$ and then lift it to the universal cover. The resulting space is isometric to $(0,\infty) \times (-\infty,+\infty)$ with coordinates $r \in (0,\infty)$, $\theta \in (-\infty,+\infty)$, equipped with the metric $dr^2 + r^2 d\theta^2$. A nonconvergent sequence $(r_n,\theta_n)$ in this space is Cauchy if and only if $r_n \to 0$: indeed the infimal path length (aka distance) from $(r_1,\theta_1)$ to $(r_2,\theta_2)$ is $\le r_1 + r_2$. For similar reasons, all nonconvergent Cauchy sequences are equivalent. Thus the completion is obtained by adding one point.
A: For a simple example, take the surface $C=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2,z>0\}$ with the metric inherited from the Euclidean metric on $\mathbb{R}^3$.  This is a cone without its vertex.  To complete it as a metric space, you would just need to add in the vertex, but this completion is not a Riemannian manifold since it is not smooth at the vertex.
