Is $\lbrace \infty \rbrace$ bounded in $\hat{\mathbb{C}}$?

Question

Is $\lbrace \infty \rbrace$ bounded in $\hat{\mathbb{C}}$? While $\hat{\mathbb{C}}$ can be identified as the riemann sphere I think you could choose $U_1(\infty) = \lbrace z \in \hat{\mathbb{C}} | \parallel z - \infty \parallel < 1\rbrace\subset \hat{\mathbb{C}}$. But $\infty - 1 = \infty$? So its kind of bounded and unbounded. I dont know. How can a set with only one point be unbounded?

Answer 1

The concept of a "bounded set" usually occurs in metric spaces $(X,d)$. See for example https://en.wikipedia.org/wiki/Bounded_set. So we have to define a metric on $\hat{\mathbb C}$. This metric generates a topology on $\hat{\mathbb C}$. In order to obtain something reasonable we must require that the subspace $\mathbb C$ of $\hat{\mathbb C}$ receives its standard topology. As you know, this is satisfied by the chordal metric $d_c$. Writing $d_c(x,y) = \lVert x -y \rVert$ is courageous because $\hat{\mathbb C}$ is not a normed linear space. However, if has a reasonable interpretation. If $S^2$ denotes the unit sphere in $\mathbb R^3$, we may identify $\hat{\mathbb C}$ with $S^2$. This is done via stereographic projection (this is a homeomorphism $h : S^2 \setminus \{(0,0,1)\} \to \mathbb R^2 = \mathbb C$ which extends via $(0,0,1) \mapsto \infty$ to a homeomorphism $H : S^2 \to \hat{\mathbb C}$). With this identification we may use the norm on $\mathbb R^3$ to get a metric on $S^2 \subset \mathbb R^3$. We simply define $d_c(x,y) = \lVert H^{-1}(x) - H^{-1}(y) \rVert$. You will see that this construction produces the chordal metric on $\hat{\mathbb C}$. See Chordal Metric - Showing it is in fact a metric.

But $S^2$ is a compact subset of $\mathbb R^3$, thus any subset of $S^2$ is bounded with respect to the norm-induced metric on $S^2$. This transfers to subsets of $\hat{\mathbb C}$ with respect to the chordal metric.