Proving compactness of the extended complex plane Prove that $(\overline{\mathbb C}, \overline{d})$ with $\overline{d}(z,z')=d(\phi(z),\phi(z'))$, where $d$ denotes the euclidean distance in $\mathbb R^3$ and $\phi$ is the inverse of the stereographic projection.
Note that $\overline{d}(w,z)=\dfrac{|w-z|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}}$, $\overline{d}(z,\infty)=\frac{2}{\sqrt{1+|z|^2}}$
I am stuck at this problem. There are a lot of equivalent ways of defining what a compact space is. The two I remember are: 1)every sequence has a convergent subsequence, 2) From any open cover of the space one can extract a finite subcover. I have no idea how to use these two definitions for $(\overline{\mathbb C}, \overline{d})$. I would appreciate suggestions and/or a definition of compactness that may be more adequate for this particular exercise.
 A: Suppose $(z_n)$ is a sequence in the extended complex plane.


*

*If there are infinitely many $n$s for which $z_n=\infty$, then you obviously have a convergent subsequence.

*If there are finitely any such $n$ but, dropping those, the sequence is unbouded, you can easily construct a subsequence which converges to $\infty$.

*Finally, if there no $n$ with $z_n=\infty$ and the sequence is bounded, you can get a convergent subsequence because closed balls in the usual plane are compact.
A: Here are the steps I would take to prove it using your second definition. Let $M$ be the unit sphere in $\mathbb{R}^3$.


*

*Prove that for every open (w.r.t. $M$) subset $S$ of $M$, there exists an open (w.r.t. $\mathbb{R}^3$) subset $\tilde{S} \subset\mathbb{R}^3$ with $S = \tilde{S} \cap M$.

*Observe from the Heine-Borel theorem that $M$ is a compact subset of $\mathbb{R}^3$.

*For any open (w.r.t. $M$) cover $C$ of $M$, use the above to show that $C$ has a finite subcover.

*Conclude that since $M$ is compact, the extended complex plane, with your distance function, is compact.

