$\mathbb{R}^n$ cannot be expressed as $A\cup B$ where $A,B$ are algebraic varieties Let $n$ be a positive integer. I want to prove one of these:
1.$\mathbb{R}^n$ cannot be expressed as $A\cup B$ where $A,B$ are algebraic varieties $V(I),V(J)$ and $A,B\neq\mathbb{R}^n$.
2.$\mathbb{R}^n$ is not Haussdorff under Zariski's topology.
3.If $V(J)=\mathbb{R}^n$ then $J=0$. 
where $I,J$ are ideals of the ring of polynomials in $n$ indeterminates.
I started trying to prove 2, then arrived to 1, which seemed very intuitive under the real numbers. But I still don't see how to prove it (I think there may be different ways to prove it outside of the algebraic geometry). Then I got to 3, but I don't think it's true in general (I'd have to use characteristic 0 I think).
Any idea or hint?
 A: To prove 3 (for $\Bbb{R}$ or any infinite field), observe that if $K$ is any integral domain and $f \in K[x]$ has infinitely many roots, then $f = 0$. Use this to show (by induction on $n$) that if $f \in K[x_1, \ldots, x_n]$ satisfies $f(\mathbf{x}) = 0$ for every $\mathbf{x} \in K^n$, then $f = 0$.
The analogue of 3 if you replace $\Bbb{R}$ by a finite field is not true. E.g., Every element of $\Bbb{Z}_2$ is a zero of $x^2-x$.
A: For I, just note that algebraic sets have empty interior for the usual topology. In particular $A\cup B$ is of empty interior.
Proof. Such a set is defined by a family of equations $f_i=0, i \in I$, where $f_i$ are not identically $0$.If it is not of empty interior, $f_i=0$ is not of empty interior. Using the Taylor formula around this interior point we see that $f_i=0$
For II, Apply I.
A: Just to provide a third perspective on the first question, you can prove the the zero set of any polynomial (so long as it is not all of $R^n$) has measure zero. 
(As a hint, use Fubini's theorem to reduce to the one dimensional case.)
