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?
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$\begingroup$ The motivation for the question is unclear, but in any case it's probably more interesting to formulate it for an arbitrary algebraically closed field of characteristic 0 (or even of prime characteristic) since this is the natural setting for the Zariski topology. Can the type of example suggested by JSpecter be adapted readily to more general fields? In other words, what role does the underlying field actually play? $\endgroup$– Jim HumphreysCommented Jun 3, 2011 at 20:04
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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.
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$\begingroup$ Second paragraph can be replaced by: Since every polynomial that vanishes on $U'$ vanishes at all points, all points are in $\bar {U'}$. $\endgroup$ Commented Jun 2, 2011 at 18:06
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$\begingroup$ In order to generalize this example to the case of an arbitrary algebraically closed field $k$ of characteristic 0, all that must be done is to redefine $U$ to be the set of pairs $(x,2^x)$ where $x\in\mathbb{Z}$. Then an identical argument as used above shows that any infinite subset $U′$ of $U$ is not contained in the vanishing of any nonzero polynomial in $\overline{\mathbb{Q}}[X,Y]$. As $U$ is contained in $\mathbb{A}^2(\mathbb{Q})$, it follows $U′$ is not contained in the vanishing of any nonzero polynomial in $k[X,Y]$. The claim then follows via Tom Goodwillie's succinct argument. $\endgroup$– JSpecterCommented Jun 4, 2011 at 9:18