Can we find a Zariski-dense subset $U$ of the affine plane over the complex number field, such that any subset $U^{\prime}$ of $U$ has its closure either the whole affine plane or finite number of points, depending the cardinal of $U^{\prime}$ is finite or not?



On Zariski Dense Subsets
Yes. Consider the set $U$ of pairs $(x,e^x)$ where $x \in \mathbb{Z}$ and let $U' $ be an infinite subset of $U.$ Then $U'$ is not contained in the vanishing of any nonzero polynomial in $\mathbb{R}[X,Y].$ As the vanishing of any polynomial $F\in\mathbb{C}[X,Y]$ is contained in the vanishing of some polynomial in $\mathbb{R}[X,Y]$, namely the product of F and its coordinatewise conjugate, it follows that $U'$ is not contained in the vanishing of any nonzero polynomial in $\mathbb{C}[X,Y].$ 


Consider the closure of $U' $ denoted by $\overline{U'} $. As a Zariski closed set, $\overline{U'} $ is the union of finitely many varieties. Choose $V$ to be an irreducible component of $\overline{U'}$ containing infinitely many points of $U'.$ The variety $V$ is infinite and therefore of dimension greater than 0. Furthermore, by our above remarks, $V$ is not the vanishing of any nonzero polynomial over $\mathbb{C}.$ As every prime ideal in $\mathbb{C}[X,Y]$ of codimension 1 is principal, it follows that $dim V \neq 1.$ We conclude $dimV = 2$ and $\mathbb{A}^2 = V = \overline{U'}.$


From this we obtain that $U$ is dense in $\mathbb{A}^2$ and the closure of any subset of U is either finite or all of $\mathbb{A}^2,$ as desired.


