Can you have a spherical plane? When I was at school I wondered if a surface could locally appear to be a unit sphere, yet `carry on forever'.  More formally, my question is:
Can you place a metric of constant curvature +1 on ${\mathbb R}^2$, such that the identity map to ${\mathbb R}^2$ (with standard Euclidean metric) is uniformly continuous?
It is possible to induce such a metric on ${\mathbb R}^2 - {\mathbb Z}^2$, by identifying each unit square with integer vertices, with a hemisphere on the unit sphere.
 A: Intrinsically, no, because one way to understand positive curvature is as a "force" that pulls diverging geodesics closer together. If the curvature is bounded from below away from zero, then any two geodesic arcs leaving a given point will intersect at a finite distance as Ulrich comments above.
But extrinsically, the situation is a lot more interesting and fun. You can embed an intrinsically flat plane in a hyperbolic 3-space and, from within the hyperbolic 3-space, it will look like it has constant positive curvature with respect to geodesic planes (intrinsically, hyperbolic planes) of the hyperbolic 3-space. There are models of the hyperbolic 3-space where the 3-space is represented by the interior of a ball of radius 1, and then one of this embedded flat planes is represented by a 2-sphere internally tangent to the surface of the 3-ball.
See http://en.wikipedia.org/wiki/Horoball, and (for a nice picture) http://en.wikipedia.org/wiki/Horocycle
A: By Myers's theorem (see http://en.wikipedia.org/wiki/Myers%27s_theorem) you have that if the Ricci curvature of a complete $n$-manifold $M$ is bounded below by $(n − 1)k > 0$, then its diameter is bounded by some constant depending on $k$. In particular, it is compact. Therefore if there is such a metric, it cannot be complete.
As pointed out by Sergei Ivanov in the comments below, the bound on the Ricci curvature holds (and therefore compactness follows) if the sectional curvature is bounded below by $k$. 
A: I don't think so:
A surface of constant positive curvature is diffeomorphic to a sphere, so it would be compact. If the identity map were continuous, then by composition the image of the surface would be a compact subset of $\mathbb R^2$ and thus not equal to the whole plane.
At least this works if the surface is orientable. If not I'm not really sure what happens (I'm a complex geometer; everything is of even dimension and orientable.)
