I want to solve Fulton's Algebraic Curves book exercise 6.5.
Any infinite subset of a plane curve $V$ is dense in $V$.
I looked for a solution here before asking and I found this: algebraic geometry exercise: infinite subset is dense. However, I didn't understand that answer (sorry if you consider this to be a duplicate) and asked my teacher for some hints. He told me to think only about irreducible curves and use the properties of Zariski topology.
This is what I thought:
$V=\mathrm V (F)$ is an algebraic set, therefore a closed set. Let $\emptyset \neq U\subseteq V$ be an infinite subset. If we find that $U$ is open, then it is dense in the Zariski topology.
I think I have to prove that $U^c$ is closed. It would be straightforward if the complementary of the infinite subset is a finite subset, because every point is a closed set and finite intersection of closed sets is closed. But I don't know if it can be that the complementary is an infinite subset too.
Let $X\subseteq V$ be the closure of an infinite subset of $V$. Then by definition of Zariski topology $X$ is the set of zeros of a finite number of polynomials. Take one of these polynomials, it defines an algebraic set which is a finite union of points and curves. Each of these curves either intersects $V$ in a finite number of points, or is equal to $V$ (this must have been shown in the previous chapter). So we see that either $X$ is finite or $X$ contains $V$, and so is dense.
See prop. 2 on page 18.