Existence of a non-variety with special properties Does there exist an infinite set $X\subseteq\mathbb{R}^n$ such that every non-zero polynomial $P\in \mathbb{R}[x_1,x_2,...,x_n]$ has finitely many zeros in $X$?
 A: Yes.  Consider the curves $(u_1(t), \ldots, u_n(t)), t \in [0,1]$, where $u_1(t) = t$ and $u_{j+1}(t) = \exp(u_j(t))$.  
Take the least $n$ for which a nontrivial polynomial $P$ in $n$ variables exists such that
 $Q(z) = P(u_1(z),\ldots,u_n(z))$ has infinitely many zeros in $[0,1]$.  Since $Q(z)$ is analytic, it is identically zero on $\mathbb C$.  Now for any $z \in \mathbb C$ and integer $k$, $u_j(z_0 + 2 k\pi i) = u_j(z_0)$ for $j \ge 2$, so $Q(z+2k \pi i) = P(z + 2 k \pi i, u_2(z), \ldots, u_n(z)) = 0$.  Since $P(s,u_2(z),\ldots,u_n(z))$ is a polynomial in $s$, that implies that $P(s, u_2(z),\ldots,u_n(z)) = 0$ for all $s$ (in particular, if $n=1$, $P$ would be identically $0$).  Choose some $s$ such that the restriction of the polynomial $P$ to $z_1 = s$ is not identically $0$, and we have $P(s, u_2(z), \ldots, u_n(z)) = P(s, e^z, u_2(e^z), \ldots, u_{n-1}(e^z)) = 0$ for all $z$, and thus $P(s, u_1(w), u_2(w),\ldots,u_{n-1}(w)) = 0$ for all $w \in \mathbb C$.  But this contradicts minimality of $n$.   
