Zariski topology of $\mathbb{C} \times \mathbb{C}$ is finer than the product topology of Zariski topologies on $\mathbb{C}$

Question

I have seen some of the posts on this topic with a slightly algebraic geometric flavour, such as Zariski topology finer than product topology. I try to understand this from a purely topological viewpoint. So the question is, is it true, that the Zariski topology on $\mathbb{C} \times \mathbb{C}$ is finer than the product of the Zariski topologies on $\mathbb{C}$?

I started with the observation that

$B = \big\{ U \subseteq \mathbb{C} \times \mathbb{C} \; \vert \; U_1, U_2 \subset \mathcal{O}^{Z}_\mathbb{C}: U = \text{pr}_{1}^{-1}(U_1) \cap \text{pr}^{-1}_{2}(U_2) \big\}$

defines a basis of the product of the Zariski topologies, which I'll denote as $\mathcal{O}_{\mathbb{C} \times \mathbb{C}}$. With $\mathcal{O}_{\mathbb{C}}^{Z}$ I denoted the Zariski topology on $\mathbb{C}$. $\text{pr}_i$ are the canonical surjections for $\mathbb{C} \times \mathbb{C}$. Let $A \subset \mathcal{O}_{\mathbb{C} \times \mathbb{C}}$ be open, then there are $U_i \subseteq B$ such that $A = \bigcup_{i \in I} U_i$. Then there are open sets $V \subseteq \mathcal{O}_{\mathbb{C}\times \mathbb{C}}^{Z}$ such that $A \subseteq V$ is open. Thus, we have that $\mathcal{O}_{\mathbb{C} \times \mathbb{C}} \subseteq \mathcal{O}^{Z}_{\mathbb{C} \times \mathbb{C}}$. Therefore, the product of the Zariski topology is finer than the product of the Zariski topologies on $\mathbb{C}$.

But I still need to show that this does not work out the other way around, such that $\mathcal{O}_{\mathbb{C} \times \mathbb{C}} \subset \mathcal{O}^{Z}_{\mathbb{C} \times \mathbb{C}}$ is a real subset. I tried an argument with the Zariski topology not being Hausdorff, but I stuck with the construction of an open set being in $\mathcal{O}^{Z}_{\mathbb{C} \times \mathbb{C}}$ but not in $\mathcal{O}_{\mathbb{C} \times \mathbb{C}}$. Thank you for your help in advance.

Answer 1

The comments of @freakish have given a false impression. The product topology makes all projections $\pi_j:\prod_iX_i\to X_j$ continuous, and is the coarsest topology with this property.

The result is that there is a sub-basis of the open sets, consisting of the “cylindrical” sets $\pi_j^{-1}U_j$ where $U_j$ is an open set of one of the factors. So, for instance, $\Bbb C\times\{w_0\}$ is closed in the product topology, as is each $\{z_0\}\times\Bbb C$. Notice that limiting the closed sets to the cofinite subsets of $\Bbb C\times\Bbb C$ does not make either projection continuous. To get the product topology from the subbase, you allow finite intersections of the cylindrical sets, and arbitrary unions of these.

For a Zariski-closed subset, take any obliqilue line $w=mz+b$ for $m\ne0$, or, much more realistically, a circle $z^2+w^2=1$.