I was wondering which fields K can be equipped with a topology such that a function f:K→K is continuous if and only if it is a polynomial function f(x)=anxn+⋯+a0. Obviously, the finite fields with the discrete topology have this property, since every function f:Fq→Fq can be written as a polynomial.

So what is with infinite fields. Does anyone see any field K where such a topology can be found? If there is no such field, can anyone supply a proof that finding such a topology is impossible. I would even be satisfied if one could prove this nonexistence for only one special field (say Q,R,C or Falgq). I suspect that there is no such topology, but I have no idea how to prove that.

(My humble ideas on the problem: Assume that you are given such a field K with a topology τ. Then for a,b∈K , a≠0, x↦ax+b is a continuous function with continuous inverse, hence a homeomorphism. Thus K is a homogenous space with doubly transitive homeomorphism group. Since τ cannot be indiscrete, there is an open set U, and x,y∈K with x∈U,y∉U. Now for every a∈K, a∗(U−y)/x includes a but not 0, and thus K∖{0}=⋃a∈K×a∗(U−y)/x, is an open subset. Thus K is a T1 space, i. e. every singleton set {x} is closed. Also K is connected: Otherwise, there would be a surjective continuous function f:K→{0,1}⊂K, which is definitely not a polynomial.)

EDIT:This question asks the analogous question with polynomials replaced by holomorphic functions. Feel free to post anything which strikes you as a remarkable property of such a hypothetical topology.

**Answer**

Currently, this is more an elongated comment than an answer …

Consider the case K=R.

Such a topology T has to be invariant under translations, scaling, and reflection because x↦ax+b with a≠0 is a homeomorphism.

*(cf. JimBelks comments above)* Assume there is a nonempty open set U bounded from below, then wlog. (by translation invariance) U⊆(0,∞) and hence (0,∞)=⋃r>0rU is open. By reflection, also (−∞,0) is open and by the pasting lemma, x↦|x|={xif x≥0−xif x≤0 is continuous, contradiction.

Therefore all nonempty open sets are unbounded from below, and by symmetry also unbounded from above.

Especially, all nonempty open sets are infinite. Let U be a nonempty open neighbourhood of 0. Let I be a standard-open interval, i.e. of the formI=(−∞,a), I=(a,∞), or I=(a,b).

Assume |U∩I|<|R| and 0∉I. Then the set {xy∣x,y∈U∩I} does not cover all of (0,1), hence for suitable c∈(0,1), the open set U∩cU is disjoint from I and nonempty (contains 0). If I itself is unbounded this contradicts the result above.

Therefore (by symmetry) |U∩(a,∞)|=|U∩(−∞,a)|=|R| for all nonempty open U and a∈R.

However, if I=(a,b) is bounded and U∩I=∅, then

⋃ca+d<acb+d>b(cU+d)=(−∞,a)∪(b,∞) is open.

Taking inverse images under a suitable cubic, one sees that all sets of the form (−∞,a)∪c,d)∪(b,∞) are open and ultimately all open neighbourhoods (in the standard topology!) of the point at infinity in the one-point compactification of R are open. A topology containing *only* these sets would describe "continuity at infinity" and make polynomials continuous - but also many other functions. Anyway we have:

|U∩(a,b)|=|R| for all nonempty U and bounded intervals (a,b), or all (standard) open neighbourhoods of ∞ are open.

Since x↦|x| is continuous under the indiscrete topology, there exists an open set ∅≠U≠R. Wlog. 0∉U. Then ⋃c>0cU=K∖{0} is open, hence

points are closed.

So T is coarser than the cofinite topology. Since x↦|x| is continuous under the cofinite topology, it is strictly coarser, i.e. there exists an open set U≠∅ such that R∖U is infinite.

**Attribution***Source : Link , Question Author : Dominik , Answer Author :
Hagen von Eitzen
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