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Came across the following statement:

Define $B_\infty(a;r)$ be the ball in $C_\infty$ with respect to the metric $d_\infty(z_1,z_2) = \frac{2|z_1-z_2|}{\sqrt{1+|z_1|^2}\sqrt{1+|z_2|^2}}$, show that if $U$ is a set in $(C_\infty,d_2)$ containing $\infty$, then $U$ is open in $(C_\infty,d_2)$ if and only if $C_\infty \setminus U$ is compact in C.

Any idea on how to prove it or Is there any book i should read on this topic? I am new to complex analysis. Thanks.

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  • $\begingroup$ What is the $d_{\infty}(z,\infty)$? $\endgroup$ – Hoseyn Heydari Jan 17 '14 at 20:29
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Give a search by stereographic projection or Riemann sphere or complex plane compactification.

Let $\phi: \mathbb{C} \longrightarrow \mathbb{S}^2\setminus N$, where $N = (0,0,1)$ ("north pole" of the unit sphere), defined by: $$\phi(z) =\frac{2}{1+|z|^2} \left(\Re z, \Im z, |z|^2 -1 \right),$$ with inverse $\psi: \mathbb{S}^2\setminus N\longrightarrow \mathbb{C} $ given by: $$\psi((u,v,t)) = \frac{1}{1-t}\left(u+iv \right).$$

The pair $(\phi, \psi)$ gives a homeomorphism (both are continuous) between $\mathbb{C}$ and $\mathbb{S}^2\setminus N$, with natural Euclidean topologies induced by the usual metrics, denoted here $d$ on $\mathbb{C}$ and $d'$ on $\mathbb{\mathbb{R}}^3$ (which the unit sphere inherits).

We now topologize $\mathbb{C}_\infty$ by adding a fundamental system of neighborhoods to $\infty$ formed by $\left\{\infty \right\}\cup \left\{z\in\mathbb{C} | |z|>r\right\}$ for all $r>0$. Note that they are mapped by extended $\phi$ (defined below) to open caps centered in $N$ on the unit sphere: $$ \phi(\left\{\infty \right\}\cup \left\{z\in\mathbb{C} | |z|>r\right\}) = \left\{(u,v,t)\in \mathbb{S}^2 | t>\frac{r^2-1}{r^2+1} \right\}.$$

We can then prove that $\psi$ extended by $\psi(N)=\infty$ and $\phi$ extended by $\phi(\infty) = \lim_{z\rightarrow \infty} \phi(z) = N$ form a homeomorphism from $\mathbb{S}^2$ to $\mathbb{C}_\infty$.

Finally, we can define $d_\infty$ as: $$ d_\infty|_\mathbb{C}(z_1,z_2) = d'(\phi(z_1), \phi(z_2)) = \frac{2d(z_1,z_2)}{\sqrt{1+|z_1|^2}\sqrt{1+|z_2|^2}} $$ and $$ d_\infty(z,\infty) = \frac{2}{\sqrt{1+|z|^2}}. $$

Note that (for non-zero complex numbers): $$ d_\infty|_\mathbb{C}(z_1,z_2) = d_\infty|_\mathbb{C}(1/z_1,1/z_2)$$ and $$ d_\infty(1/z,\infty) = \frac{2z}{\sqrt{1+|z|^2}}. $$

It can be shown that $\left(\mathbb{C}_\infty, d_\infty\right)$ is a metric space. Moreover, it is compact as it is the image of a compact, $\mathbb{S}^2$ (bounded and closed in $\mathbb{\mathbb{R}}^3$), through a continuos function $\psi$.

To answer your question the definition of the compact metric space $\left(\mathbb{C}_\infty, d_\infty\right)$ should help or, using earlier observations, an open set around $\infty$ is mapped to an open set around $N$ (homeomorphism is an open map) whose complement is closed and compact (as it is bounded) which gets mapped back in the complement of the open set (we started with) that must be compact (image of a compact through a continuous function).

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