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?

  • $\begingroup$ 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 $\mathbb{C}$). $\endgroup$
    – user512346
    Commented Nov 25, 2018 at 1:25
  • $\begingroup$ 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. $\endgroup$
    – Somos
    Commented Nov 25, 2018 at 2:15
  • $\begingroup$ 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. $\sup_\sigma |a^\sigma|$. $\endgroup$
    – reuns
    Commented Nov 25, 2018 at 2:46


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