Is an open disk complete?

Question

See the definitionS of complete surface, first definition: without edges. second definition: Any line segment can be continued indefinitely. By the first, open disk seems to be a complete surface, but by the second, it seems not. So if it is complete on earth? Thank you.

Answer 1

It's not clear what you mean by complete. Here are two possible definitions:

1) It is complete as a metric space: every Cauchy sequence converges to a point in the space.

2) It is geodesically complete - i.e. at each point the exponential map is defined on the whole tangent space at that point.

(By Hopf-Rinow these two are equivalent.)

However, these definitions don't make sense without a metric. In the former case any metric in the sense of a metric space will do, but for the latter to make sense you probably want a Riemannian metric. For concreteness let's say we take the metric-space notion for this discussion.

Here are two examples two show why this matters (both will be Riemannian metrics).

Example 1: The open disc is homeomorphic to R^n, so you may take the standard Euclidean metric on R^n and pull it back to the disc, so that the two are isometric. With this metric it is complete because it is isometric to a complete space (R^n).

Example 2: Take the Euclidean metric in R^n and just restrict it to the disc (i.e. the usual metric). This space is not complete as above because you may take as a Cauchy sequence any sequence converging to a boundary point, and this sequence fails to converge in the disc. (Note that in the above example this sequence would not be Cauchy because the metric in that example blows up near the edge of the disc.) This shows that this metric on the disc is not complete.

You could take as another example the Poincare (hyperbolic) metric on the disc, which is quite different from example 1 (negatively curved) but is also complete.