How can the meromorphic functions be the rational ones?

I am reading Stein's book on Complex Analysis, and it defines:

Definition: A function $$f$$ on an open set $$\Omega$$ is meromorphic if there exists a sequence of points $$\{z_0, z_1, z_2, ... \}$$ that has no limit point in $$\Omega$$, and such that

$$(i)$$ the function $$f$$ is holomorphic on $$\Omega - \{z_0, z_1, z_2, ... \}$$

$$(ii)$$ $$f$$ has poles at the points $$\{z_0, z_1, z_2, ... \}$$

A bit later, there is this theorem

Theorem: The meromorphic functions in the extended complex plane are the rational ones.

From the definition, it appears that meromorphic functions can have an infinite number of discontinuities. However, I believe rational functions are of the form $$\dfrac {p(z)}{q(z)}$$ where $$p,q$$ are polynomials, so they can have a maximum of $$\deg q$$ number of poles. How can a function with an infinite number of poles be rational?

• The extended complex plane is compact, so there is no infinite sequence with no limit point on it – kneidell Mar 3 at 17:12
• @kneidell Ah okay thanks! – Ovi Mar 3 at 17:14