TITLE: Is $\mathbb{C}^2$ homeomorphic to $\mathbb{C}^2 - (0,0)$ with the Zariski topology?
QUESTION [28 upvotes]: A fellow grad student asked me this, I have been playing for a while but have not come up with anything.  Note that $\mathbb{C}$ is homeomorphic to $\mathbb{C} - \{0\}$ in the Zariski topology - just take any bijection and the closed sets (finite sets) will biject as well.  Concocting a similar thing for the plane is harder though.
I think I can show that the rational plane and the rational plane minus the origin are homeomorphic by enumerating the irreducible curves and using a back and forth argument, but I have not written it all up formally to see if I am missing something yet.
I know the question isn't natural from the point of view of algebraic geometry, because one of the objects isn't even a variety.  I think it is still interesting just to see how weird the zariski topology really is.

REPLY [15 votes]: This question was analyzed by  Roger Wiegand  in the context of algebraic surfaces over fields $k$ which are algebraic closures of finite fields. 
[1] R. Wiegand, Homeomorphisms of affine surfaces over a finite field, J. London Math. Soc. (2) 18 (1978), no. 1, 28–32.
and in the followup paper:
[2] R. Wiegand, W. Krauter, Projective surfaces over a finite field. Proc. Amer. Math. Soc. 83 (1981), no. 2, 233–237.
In [1] he proves that the affine plane $A^2_k$ (over $k$) is Zariski-homeomorphic to any open nonempty subset of $A^2_k$ (Corollary 7). In particular, $A^2_k$ is homeomorphic to $A^2_k -\{(0,0)\}$. In particular, Greg's argument (as written) cannot be complete (as zero characteristic is not used anywhere in his sketch). 
After proving the corollary Wiegand asks if the same holds for $k={\mathbb R}$ and ${\mathbb C}$. 
In [2] it is proven (amon other things) that every proper nonempty open subset of $P^2_k$ is homeomorphic to either $P^2_k - \{point\}$, or to the affine plane $A^2_k$.