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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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1 Answer 1

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.)

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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

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