Show that the closed unit ball in \(\mathbb{R}^2\) is a 2-manifold with boundary.
Solution
Let \[ D^2=\{(x,y)\in\mathbb R^2:x^2+y^2\le1\}. \] It is Hausdorff and second countable as a subspace of \(\mathbb R^2\).
If \(p\in D^2\) satisfies \(\|p\|<1\), choose \(\varepsilon>0\) with \(B(p,\varepsilon)\subset D^2\). This is an open neighborhood of \(p\) in \(D^2\) homeomorphic to an open subset of \(\mathbb R^2\).
Now let \(p\in\partial D^2=S^1\). After a rotation we may assume \(p=(1,0)\). For sufficiently small \(\varepsilon\), the neighborhood \(D^2\cap B(p,\varepsilon)\) is carried homeomorphically to an open subset of the closed half-plane \[ H^2=\{(u,v):v\ge0\}. \] One explicit chart is obtained by polar coordinates near \(p\): write a nearby point uniquely as \[ (r\cos\theta,r\sin\theta), \] with \(r\le1\) and \(|\theta|<\delta\), and send it to \[ (\theta,1-r). \] For sufficiently small \(\delta\) this is a homeomorphism onto an open subset of \(H^2\), with \(p\mapsto(0,0)\).
Thus every point has a neighborhood modeled on \(\mathbb R^2\) or \(H^2\), so \(D^2\) is a 2-manifold with boundary. Its boundary in this sense is the usual circle \(S^1\).