# Find Metric in $\mathbb{R}^2$ s.t. it is not Complete

My friend ask me: How to define a metric in $\mathbb{R}^2$ in such an way that $\mathbb{R}^2$ is not complete. I gave him the following metric:

Let $B=\{x\in\mathbb{R}^2:\ \|x\|<1\}$. By a diffeomorphism we can think that $\mathbb{R}^2$ is $B$. In this way we have that the points in $B$, close to the boundary of $B$, are the points in $\mathbb{R}^2$ with big norm in $\mathbb{R}^2$. Hence, if $F:\mathbb{R}^2\rightarrow B$ is the diffeomorphism, we can define the metric in $\mathbb{R}^2$ by $$d(x,y)=\overline{d}(F(x),F(y))$$

where $\overline{d}$ is the euclidean metric restricted to $B$.

He liked the metric, but he asked me an more "elementary" metric, not so trivial but not so elaborated.

Thanks

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Maybe it looks more elementary if you give the map explicitly. Or try $d((x_1,y_1), (x_2,y_2)) = |y_1-y_2|+|\arctan x_1-\arctan x_2|$. And of course instead of diffeomorphism, any map will work, e.g. a bijection with $\mathbb R\setminus\mathbb Q$. –  Hagen von Eitzen Oct 29 '12 at 21:03

The standard example is stereographic projection to the standard unit sphere, through the North Pole. You are not saying that you have background in differential geometry, so we will give a recipe: the north pole is $(0,0,1)$ in the sphere $x^2 + y^2 + z^2 = 1.$ For any point $P$ in the $xy$-plane, draw the line segment that joins $P =(x,y,0)$ to $(0,0,1).$ The segment intersects the sphere in a second point, call that $P^\ast.$

Next, if I have two points on the sphere, there is a distance between them given by drawing the arc of a great circle that passes through both, but is the shorter of the two pieces of this great circle. This length is never longer than $\pi.$

So that is your metric, for two points $P,Q$ in the plane, find $P^\ast,Q^\ast.$ The great circle distance between those is defined to be the new distance between $P,Q.$

This metric is incomplete because nothing maps to the North Pole itself.

Note that there are different conventions about where to put the sphere. The one I give is closely tied to the use of the $xy$-plane as the complex numbers $\mathbb C.$ The sphere is often called the Riemann Sphere, and the whole business is tied to Möbius transformations. In this version, the standard unit circle in $\mathbb R^2$ is identified neatly with the standard unit circle in $\mathbb C.$ It's all a matter of choice.

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Thanks yoy Will. I gonna accept your answer, because "for me" its looks like the most simple way to define the metric (compared with the other answers). It is more geometric. –  Tomás Oct 30 '12 at 0:08
What does it mean to be complete? My understanding was that all Cauchy sequences converge? so how is this equivalent to saying "its incomplete because nothing maps to the North Pole itself" Thanks –  Tyler Hilton Oct 11 '13 at 0:55
@TylerHilton, take a sequence of points on the sphere converging to the north pole. As we have deleted the north pole from the space, this is a Cauchy sequence that does not converge. Its image in $\mathbb R^2$ after stereographic projection is also Cauchy sequence that does not converge. The really quick version is this: in $\mathbb R^2,$ the sequence of points $p_n = (n,0)$ is now a Cauchy sequence under the revised metric space structure. –  Will Jagy Oct 11 '13 at 1:14

Since everybody tries something almost everywhere smooth, I repeat my non-smooth solution from the comment above as an asnwer:

The set $\mathbb R^2$ as the same cardinality as $\mathbb R\setminus \mathbb Q$. Therefore there exists a bijection $F\colon \mathbb R^2\to \mathbb R\setminus \mathbb Q$. Then we can define the metric $$d(x,y)=|F(x)-F(y)|$$ on $\mathbb R^2$, which makes it an incomplete metric space, of course isomorphic to $\mathbb R\setminus \mathbb Q$ with standard metric.

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You are a man of stern principle. –  Will Jagy Oct 29 '12 at 21:25
Very nice example, Thanks. –  Tomás Oct 30 '12 at 0:04

While the motivation for your metric is a little bit elaborate, finding an actual diffeomorphism is easy and makes it almost as easy to find an explicit form for your metric $d$. Another version of your construction that's about as easy to compute with: consider the classic stereographic projection from $S^2\to\mathbb{R}^2$ given by treating the $S^2$ as a unit sphere centered at $(0,0,1)$ and using the intersection of the line between the 'north pole' $p=(0,0,2)$ and the $xy$-plane to map each point in $\mathbb{R}^2$ to a point in $S^2-p$. You can then parametrize the sphere via longitude and latitude $(\theta, \phi)$ and use those to compute the metric; in fact, you should even be able to use the (non-uniform) metric $d(a,b) = (a_\theta-b_\theta)^2+(a_\phi-b_\phi)^2$ and project this out to get an incomplete metric in terms of polar coordinates on $\mathbb{R}^2$.

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Thanks you Steven –  Tomás Oct 30 '12 at 0:06