= Solution
Use <countability of negative curves>. A <smooth projective surface> over $\mathbb C$ is a <compact> <smooth manifold> homotopy equivalent to a <finite CW complex>, so its integral <cohomology group> $H^2(X,\mathbb Z)$ is finitely generated and therefore countable. Distinct irreducible curves $C,D$ with $C^2<0$ cannot have the same <cohomology class>: that would give $C\cdot D=C^2<0$, while distinct effective curves have nonnegative <intersection multiplicities>. Assigning a <negative curve> its <cohomology class> is thus <injective>, proving \b[there are at most countably many negative curves].
For zero <self-intersection>, uncountable families do occur. On the product of two <projective lines> $\mathbb P^1\times\mathbb P^1$, every fiber $\mathbb P^1\times\{p\}$ has <self-intersection> zero, since a different fiber is disjoint and has the same <numerical equivalence of divisors> class. The parameter $p\in\mathbb P^1(\mathbb C)$ varies over an uncountable set. Thus \b[yes, there can be uncountably many such curves].
Back to article page
