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Let $\hat{\mathbb{C}} = \mathbb{C} \cup \{ \infty \}$ denote the Riemann Sphere. While reading the wikipedia article about it I found a passage that said that every rational function on the complex plane can be extended to a continuous function on the Riemann Sphere.

The particular construction is as follows, let $R(z) = \frac{f(z)}{g(z)} \in \mathbb{C}(z)$ be a rational function. Lets assume for simplicity that $f(z)$ and $g(z)$ share no common factor. Then for any point $a \in \mathbb{C}$ such that $g(a) = 0$ but $f(a) \neq 0$ we define $R(a) = \infty$. Also we define $R(\infty) := \lim_{z \to \infty} R(z)$.

So to quote wikipedia, with this definitions $R(z)$ becomes a continuous function from the Riemann Sphere to itself. The problem for me is that the article doesn't add any details as to how one may go about showing that in fact $R(z)$ is continuous. So my question is exactly that, to make it simple, if I have for instance $R(z) = \frac{z-1}{z+1}$ or even $R(z) = \frac{1}{z}$, how do I show that it is a continuous function on $\hat{\mathbb{C}}$? I think I can build up from an easy example such as this (assumming this is easy).

Also, I'm a little bit confused about how to interpret this continuity, how should I see the continuity on the Riemann Sphere?, does it involve an argument with stereographic projection?

I added in the Riemann Surface tag just in case they are involved, which I'm not sure. Thank you very much in advance.

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This question depends quite a bit on what precisely you are taking as your definition of $\mathbb{C}\cup\{\infty\}$. In particular, you've told us what it is as a set, but what is the topology on it? You could use the "one-point compactification", or you could think of it as having the subspace topology when embedded in $\mathbb{R}^3$, or probably others. They are equivalent, but until you pick one specifically it is hard to show something is continuous. –  Matt Nov 21 '10 at 17:47
    
Well I'm not sure since the article in wikipedia didn't mention which one in particular is the one that works. If it is the case that both topologies do work, then I would prefer the one which makes it easier to show the continuity of the rational functions. –  Adrián Barquero Nov 21 '10 at 17:53

2 Answers 2

up vote 3 down vote accepted

One way of doing this is to use the sequential definition of continuity... this works because of the way the Riemann sphere has a metric (distance function) defined on it, inherited from the regular Euclidean distance function in 3-d. In this viewpoint, the statement that $\lim_{z \rightarrow a} f(z) = \infty$ is equivalent to saying $\lim_{z \rightarrow a} |f(z)| = \infty$. In the case that $a$ itself is infinity, then $\lim_{z \rightarrow \infty} f(z) = \infty$ becomes $\lim_{|z| \rightarrow \infty } |f(z)| = \infty$. In other words, for every $N$ there's an $M$ such that if $|z| > M$ then $|f(z)| > N$. Similarly, for some finite $z_0$ the statement $\lim_{z \rightarrow \infty} f(z) = z_0$ means $\lim_{|z| \rightarrow \infty } f(z) = z_0$; for every $\epsilon > 0$ there's an $N$ such that $|z| > N$ implies $|f(z) - z_0| < \epsilon$.

In this way many statements like the ones you're trying to prove become pretty routine.

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I suppose you're talking about the so called "chordal metric" on the Riemann Sphere? –  Adrián Barquero Nov 21 '10 at 18:10
    
yep, that's the one. –  Zarrax Nov 21 '10 at 18:25
    
Ok, thanks. I'll try with this. –  Adrián Barquero Nov 21 '10 at 18:31
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It does not really matter what metric you take, as long as it is compatible with the topology given by the local charts. Actually one needs only to perform the coordinate change $z\mapsto 1/z$ so that the point $\infty$ has the coordinate $0$. –  timur Dec 24 '10 at 6:04

As a lowly high school teacher, my response will probably appear to be comical; nonetheless, I've always imagined rational functions to be continuous. As $x\to\infty$, I imagine the $x$-axis as a great circle of a unit sphere. The $y$-axis would also be a great circle. In spherical geometry, two lines intersect at two poles. If the origin is located at one pole, and infinity at the other, the function can be envisioned as being continuous at infinity. As timur mentions, then infinity would have coordinate $0$.

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