Two easy questions about algebraic varieties I'm studying Fraleigh's abstract algebra(7ed), and there are little contents about algebraic geometry, just the definitions of varieties and ideals. Since I have few backgrounds about algebraic geometry, I don't know how to solve the following two exercises. 

Sec28, Ex27, b. Give an example of a subset of $\mathbb{R}^2$ which is  not an algebraic variety.
Ex34. Give an example of a subset $S$ of $\mathbb{R}^2$ such that $V(I(S))\neq S$.
  (Here, the algebraic variety $V(S)$ in $F^n$ is the set of all common zeros in $F^n$ of the polynomial in $S$, where $S$ is a finite subset of $F[\mathbf{x}]$.)

I think that the answer of the two exercises can be same. But I don't know how to show some subset is not an algebraic variety. How can I solve it?
 A: Hint1: A univariate polynomial can have infinitely many zeros only if it is the zero polynomial.
Hint2: Show that if the polynomial $p(x,y)$ has infinitely many zeros on the line $y=y_0$, it must be divisible by $y-y_0$.
Hint3: Show that if the conclusion of the previous hint holds for infinitely many choices of $y_0$, then $p(x,y)$ must be the zero polynomial.
A: The easiest solution to the first exercise depends on exactly what your running definition of a subvariety of $\mathbb{R}^2$ is.  But no one would call a subset of affine $n$-space a subvariety unless it is at least locally Zariski-closed, i.e., the intersection of a Zariski-closed subset and a Zariski-open subset.  It is not too  hard to show that the complement of any countably infinite subset of $\mathbb{R}^2$ is not of this form.  One could for instance proceed as follows:


*

*Show that any proper Zariski-closed subset of $\mathbb{R}^2$ has measure zero.  

*Show that any infinite Zariski-closed subset of $\mathbb{R}^2$ is uncountable.  


[On second thought, maybe 2. is overkill here.  If you take your countably infinite subset to be contained in, for instance, the line $y = x$, then it is much easier than what I proposed above to see that it is not Zariski-closed, so that its complement is not Zariski-open.]
As for the second exercise, you can take any non-Zariski closed subset, so for instance any subset which is not closed in the Euclidean topology.  As you say, any solution to the first exercise will certainly do here.  
