Does a dense subset of a projective variety has the same dimension as the variety

Question

I know that an open subset of a projective variety, which is dense, has the same Krull dimension as the projective variety and I also know that, in general, a dense subset may not have the same Krull dimension as the set. Therefore I wonder if a dense subset of a projective variety will have the same Krull dimension? Here, the projective variety is the standard closed subset of projective space under Zariski topology and Krull dimension means the supremum of lengths of chains of closed irreducible sets.

Answer 1

$$S=\{ (1/n,e^{-n}),n\ge 1\}\subset \Bbb{A^2(C)}$$ $S\cap V(f)$ is finite for any non-zero polynomial $f\in \Bbb{C}[x,y]$.

(A parametrization of a locally irreducible analytic hypersurface passing through $(0,0)$ is of the form $(t^k,g(t))$ or $(g(t),t^k)$ with $g(0)=0$ and $g$ analytic near $0$, so it can't pass through infinitely many points of $S$)

Thus $S$ is dense in $\Bbb{A^2(C)}$.

$\Bbb{A^2(C)}$ has a chain of 3 irreducibles subsets whereas, since $S\cap V(f)$ is finite, $S$ has a chain of only 2

$$(0,0)\subset V(x)\subset \Bbb{A^2(C)},\qquad (1/n,e^{-n})\subset S$$

Replacing $\Bbb{A^2(C)}$ by $\Bbb{P^2(C)}$ answers your question.