Another abstract algebra/field theory question 
Suppose that $F$ is a field, $S \subseteq F^n$ and $I$ is an ideal in $F[x_1, \cdots, x_n] = F[\bar{x}]$. Define
  $$I(S) = \{ f \in F[\bar{x}]: f(\bar{s}) = 0, \forall \bar{s} \in S\}$$ and
  $$V(I) = \{\bar{s} \in F^n: f(\bar{s}) = 0, \forall f \in I \} $$
  Give an example of a subset $S$ of $\mathbb{R}^2$ for which $S \neq V(I(S))$.

I know that $I(S)$ is an ideal and I also know that $S \subseteq V(I(S))$.
I cannot think of any subsets of $\mathbb{R}^2$ that satisfy this. Any hints?

For an easy approach, there exist sets $S\subsetneq \mathbb{R}^2$ such that a polynomial $f\in \mathbb{R}[x,y]$ that vanishes at every point in $S$ must be identically zero, for example $S=\mathbb{R}^2\setminus\{\overline{s}\}$ for any point $\overline{s}=(s_1,s_2)$ (because a polynomial function is continuous in the standard topology on $\mathbb{R}^2$). Then we have $$S\neq \mathbb{R}^2=V(I(S)).$$
