# Why euclidean topology on complex numbers?

This is a rather inconcrete question and i am hoping for different answers:

The topology on $$\mathbb{Q}$$ and $$\mathbb{R}$$ is natural in being the order topology of these ordered fields.

Endowing $$\mathbb{C}$$ with the topology of $$\mathbb{R}^2$$ is the key to all those nice results of classical complex analysis and the important fact, that a function $$\mathbb{C} \to \mathbb{C}$$ is differentiable if and only if it is a conformal (locally angle-preserving) map. So the theory of holomorphic functions is more on geometry of $$\mathbb{R}^2$$ than on the properties of $$\mathbb{C}$$ as a field, I think.

Now suppose, the existence of an algebraic closure of $$\mathbb{Q}$$ would have been discovered before the invention of $$\mathbb{C}$$. It is hard to imagine that someone was like "lets endow $$\mathbb{Q}[\sqrt{-1}]$$ with the product metric of $$\mathbb{Q}^2$$ and embed its algebraic closure into its metric completion!"

Because there are so many other choices: Choose $$\sqrt{2}$$ instead of $$\sqrt{-1}$$; endow $$\mathbb{Q}[\sqrt{-1},\sqrt{2}]$$ with the product topology of $$\mathbb{Q}^3$$; etc.

My question is: Why does $$\mathbb{C}$$ deserve the euclidean topology of $$\mathbb{R}^2$$ and are there other choices?

• In fact, there are other choices, depending on what you're interested about $\mathbb{C}$. For instance, in Algebraic Geometry one does not endow $\mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $\mathbb{C}$).
– user512346
Commented Nov 25, 2018 at 1:25
• You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices. Commented Nov 25, 2018 at 2:15
• What is the field $\mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $\mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $\mathbb{Q}(b_1,b_2,\ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $\sup_\sigma |a^\sigma|$. Commented Nov 25, 2018 at 2:46