Topology of product of affine varieties In this wiki article: https://en.wikipedia.org/wiki/Affine_variety#Products_of_affine_varieties, it states that polynomials that are in $k[x_1, \ldots, x_n, y_1, \ldots, y_m]$ but not in $k[x_1, \ldots, x_n]$ or $k[y_1, \ldots, y_m]$ define algebraic varieties that are in the Zariski topology of $\mathbb{A}^n \times \mathbb{A}^m$ but not in the product topology. However, $\mathbb{V}(x_1y_1)$ satisfies this condition, yet is in the product topology: $\mathbb{V}(x_1) \times \mathbb{A}^m \cup \mathbb{A}^m \times \mathbb{V}(y_1).$ Any guidance would be appreciated.
Furthermore, is it true that for any nonfinite affine varities $X, Y$ that the Zariski topology $X \times Y$ is not the product topology? 
 A: That claim is incorrect: rather, it should be that any polynomial in $k[x_1,\cdots,x_n,y_1,\cdots,y_m]$ which can't be written as a product of polynomials in $k[x_1,\cdots,x_n]$ and $k[y_1,\cdots,y_m]$ defines a subvariety not closed in the product topology.
Your guess needs a little refinement in order to work: if you're working over a field $k$ and your product is really $\times_k$, this is true. Embed $X$ and $Y$ in to $\Bbb A^n_k$ and $\Bbb A^m_k$ respectively, which induces an embedding $X\times_k Y \to \Bbb A^n_k\times_k \Bbb A^n_k$. Pick some subvariety $Z$ of $\Bbb A^{n+m}_k$ which is not closed in the product topology and intersects $X\times_k Y$ with dimension at least 1. Then $Z\cap X\times_k Y$ cannot be a closed subvariety of the product topology on $X\times_Y K$.
(Note that this is false if we work over the integers and $X=Y=\operatorname{Spec}\Bbb Z$. The product here is over $\operatorname{Spec} \Bbb Z$ so $X\times Y = X =Y$, which is definitely affine, has the product topology, and is not finite. It's always good to specify your base!)
