# If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $A^1 \times A^1$ zariski topology.

I am not sure what is incorrect about the statement.

If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $$A^1 \times A^1$$ zariski topology, i.e singletons should be closed.

• (let us assume the base field is algebraically closed) whenever $$a$$ and $$b$$ are closed points in $$\mathbb{A}^1$$, $$(a,b)$$ is indeed closed in $$\mathbb{A}^2=\mathbb{A}^1\times \mathbb{A}^1$$
• the Zariski topology on $$\mathbb{A^2}$$ is not the product topology, it is finer. For example $$\{y-x^2=0\}$$ is not closed under the product topology but it is under the Zariski topology.
• Proper closed subsets of $\mathbb{A}^1$ are finite, so closed subsets of the product are also finite, or of the form $\mathbb{A^1}\times \mathrm{finite}$. $\{y-x^2\}$ is not. Apr 18, 2023 at 9:13
• @benhuni I don't understand what you are saying. What is not true? What do you mean by "topologically not true"? Homeomorphism between what exactly? And why does it matter? Even if the solution set is homeomorphic to $A^1$, so what? This is a different, unrelated question. Apr 18, 2023 at 11:44
• @benhuni one last time: homeomorphism between what? Existence of some random homeomorphism has nothing to do with the question whether a subset is closed or not. For example $(0,1)$ is homeomorphic to $\mathbb{R}$ but it doesn't make it closed in $\mathbb{R}$. Again: these are unrelated things. There's nothing wrong with the answer. Apr 18, 2023 at 15:31