Zariski Topology question

Just to add to Prof Magidin's example: it's a common exercise in point set topology to prove that a topological space $X$ is Hausdorff if and only if the diagonal $\Delta = \{(x, x) : x \in X\}$ is closed in the product topology on $X \times X$. And when $k$ is infinite [in particular, when $k$ is algebraically closed] you know that $\mathbf A^1$ is not Hausdorff: any two non-empty open sets intersect.

By the way, this consideration of the diagonal motivates the correct analogue of the Hausdorff property in algebraic geometry: the notion of a separated morphism of schemes.

Assuming $k$ is infinite, consider the topological properties of the line $x=y$.

Zariski closed sets in $\mathbb{A}^1$ are sets of finitely many points.

So, open sets are compliments of finitely many points.

Let $X,Y$ be topological spaces, then product topology on $X\times Y$ is topology generated by basis elements of the form $U\times V$ where $U$ is open in $X$ and $V$ is open in $Y$.

So, product topology on $\mathbb{A}^1\times \mathbb{A}^1$ is generated by $$\mathbb{A}^1\setminus\{a_1\cdots,a_r\}\times \mathbb{A}^1\setminus\{b_1,\cdots,b_l\}=\mathbb{A}^1\times \mathbb{A}^1\setminus\{(a_i,b_j):1\leq i\leq r;1\leq j\leq l\}$$

Topology generated by basis elements $\{U_{\alpha}\}$ is countable unions of elements of the collection $\{U_{\alpha}\}$

So, any open set in product topology is countable union of sets of the form $$\mathbb{A}^1\times \mathbb{A}^1\setminus\{(a_i,b_j):1\leq i\leq r;1\leq j\leq l\}$$

Countable union of finitely many points is countable.

So, any open set in product topology is complement of countably many points.

So, any closed set in product topology is set of countably many points.

Assuming we are in $\mathbb{C}$, $\{(x,y):x=y\}$ is an infinite uncountable set, closed in Zariski topology, thus not a closed set in product topology.

Thus, product topology is not the same as zariski topology.