Differential Geometry: Is a closed disk a surface?

Question

An open disk is clearly a surface, in the sense that it is locally homeomorphic to a part of $\mathbb{R}^2$. But what about a closed disk, even though it still looks like a surface, I am starting to think that the points on the boundary will pose problem? I feel like I'm asking a silly question here, am I misunderstanding something fundamental here?

Answer 1

One usually distinguishes manifolds from manifolds with boundary; in the case of a manifold with boundary, the thing to check is that is locally homeomorphic to a neighborhood of the upper halfplane $\mathbb{H} = \{(x,y) \in \mathbb{R}^2 : y \geq 0\}$.

Answer 2

Here's an outline of a proof that the closed unit disk is not a continuous manifold.

If $x$ is in the boundary of the closed disk, and $U$ is a contractible open neighborhood of $x$, then $U-\{x\}$ is again contractible. As $\Bbb R^2-\{pt\}$ deformation retracts to $S^1$ (which is not contractible), $U-\{x\}$ cannot be homeomorphic to $\Bbb R^2-\{pt\}$, and therefore $U$ is not homeomorphic to $\Bbb R^2$. (This extends to higher dimensional disks if you believe $S^n$ is never contractible.)