# Continuity in the extended complex plane

Ahlfors says that for rational function $R(x) = \frac{P(z)}{Q(z)}$, we define $R(z)$ to be $\infty$ when $Q(z) = 0$. Then he says that $R(z)$ is clearly continuous.

To me, $R(z)$ is clearly continuous at the points where $Q(z) \not = 0$. But for continuity, you need $|R(z) - \infty| = \infty < \epsilon$ for $z$ which are close to the zero. So $R(z)$ can't be continuous in the sense that it is continuous at every point.

Is Ahlfors being informal here, or am I missing something fundamental?

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First off, you need the additional assumption that $P(z)$ is nonzero where $Q(z)=0$ to make sense of this statement. Otherwise, all bets are off when you take the limit.
The usual $\epsilon$-$\delta$ definition doesn't apply to inifinity. Instead, we say that $\lim_{z \to z_0}f(z)=\infty$ if for all $M\in \mathbb{R}$ there exists an $\delta$ so that whenever $|z-z_0|<\delta$, $|f(z)|>M$. It now makes sense to say the $R$ is continuous at $z$ in the sense that the value for the function is equal to the limit as the independent variable approaches $z$.
(Notice that in real analysis, we make a distinction between positive and negative $\infty$, but in complex analysis we just work with a single infinite limit point. A good visual for this is the Riemann Sphere, which you will no doubt encounter as you continue reading Ahlfors.)
Note that this is not an ad hoc definition just for infinite limits; it's the application of the standard topological definition of a limit to the standard topology for the real numbers extended by $\pm\infty$ or the complex numbers extended by $\infty$. In addition to the open sets of unextended numbers, that topology has open sets containing $\infty$, namely the complements of compact sets. The above definition says precisely that the function is outside any compact set and thus inside any neighbourhood of $\infty$ for sufficiently small $\delta$. –  joriki Mar 27 '12 at 16:35