Naive algebraic geometry question Please no flame, I am a beginner here... Given ANY subset of $\mathbb{C}$, is there always at least one set of polynomial equations which generates it? If yes, can we always explicitly construct at least one member of that set (of polynomial equations)?
What is the situation for $\mathbb{R}^2$ and polynomials in $x,y$? For example, $x^{2}+y^{2} = 1$ generates the unit circle in $\mathbb{R}^2$.
 A: The question is a little hard to interpret, I assume that by a set of polynomials generating a subset of $\mathbb C$ you mean that the subset of $\mathbb C$ is the set of common zeros of the polynomials in your set.  Of course this means you should only be considering polynomials in a single variable, not in multiple variables as you explicitly stated.  So maybe I've misinterpreted you, let me know if I have.
Now assuming I have interpreted your question correctly, what you are asking is whether the Zariski topology on $\mathbb C$ is discrete.  The answer is no, the Zariski topology on $\mathbb C$ is the cofinite topology.  This means that the closed sets (meaning the sets that arise as the common zeros of a set of polynomials) are exactly the finite subsets of $\mathbb C$ and $\mathbb C$ itself.
To see this first note that $\mathbb C$ is the solution set of the set of polynomials $\{0\}$.  For any other set $S$, if $f \in S$ is any nonzero polynomial then $f$ has only finitely many solutions and the common zeros of $S$ must be a subset of this finite set.  This shows that every closed proper set is finite.  To see that every finite set is closed let $\{a_1, \ldots, a_n\}$ be a finite set of points in $\mathbb C$.  Then the zeros of the set $\{f\}$, where $f = (x - a_1)\cdots(x - a_n)$, are exactly the $a_i$.
Edit: For $\mathbb R^2$ the answer is still no, not every subset of $\mathbb R^2$ is the solution set of a set of polynomials.  The standard example is the set $\mathbb R^2 \setminus (0, 0)$.  I claim that the zero polynomial is the only polynomial that is zero on this set.  To see this let $f(x, y) = \sum_{i = 0}^ng_i(x)y^i$ be any polynomial that is zero on $\mathbb R^2 \setminus (0, 0)$.  Then for any $a \neq 0$ the polynomial $f(a, y)$ is identically zero.  This means each $g_i(x)$ is a polynomial with the property that $g_i(a) = 0$ when $a \neq 0$.  As polynomials give continuous functions we then get $g_i(0) = 0$ so $g_i$ is the zero polynomial.  Then $f$ is also the zero polynomial.
So we see that no set of polynomials has $\mathbb R^2 \setminus (0, 0)$ as their zeros because those polynomials are zero, and hence give the set $\mathbb R^2$ and not $\mathbb R^2 \setminus (0, 0)$.
More generally, any polynomial that's zero on an open set is zero, so you can't get any subset that contains an open set (open in the Euclidean topology).  Also, I don't believe you can get a proper subset of positive Lebesgue measure, but you certainly can't get all subsets of Lebesgue measure $0$ so that's still not a complete characterization.
A: Adapted from my comment:
For the revised question about $\mathbb{R}^2$, the answer is still no.
First, a note: every polynomial equation $f=g$ is equivalent to $f-g=0$, thus instead of talking of the locus of solutions to a set of polynomial equations, we may talk about the locus of zeros of a set of polynomials.
Let $X=\mathbb{R}^2-\{(0,0)\}$. 
Suppose for contradiction $X$ was described as the locus of zeros of a set $S$ of polynomials. If $S$ consists only of the polynomial $0$, then the locus of zeros is all of $\mathbb{R}^2$, so suppose there is a nonzero polynomial $f\in S$. Now, $f$ is a polynomial potentially in both $x$ and $y$.
Case $1$: $f$ has terms involving $x$. Then if we do the substitution $y=1$, we get a nonzero polynomial, $f_x$, which involves only $x$. Since this is a one-variable nonzero polynomial, it has only finitely many zeros. So let $a$ be a value which is not a zero of $f_x$. We see that $(a,1)$ is not a zero of $f$.
Case $2$: $f$ has terms involving $y$. Similarly to above, we can find a point $(1,b)$ which is not a zero of $f$.
In conclusion, there is a point other than the origin which is not a zero of $f$. So the locus of points where the set $S$ of polynomials vanishes is necessarily smaller than $\mathbb{R}^2-\{(0,0)\}$.
