A Topology such that the continuous functions are exactly the polynomials

I was wondering which fields K can be equipped with a topology such that a function f:KK 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:FqFq 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,bK , a0, xax+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,yK with xU,yU. Now for every aK, a(Uy)/x includes a but not 0, and thus K{0}=aK×a(Uy)/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.


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 xax+b with a0 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 x0xif x0 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 |UI|<|R| and 0I. Then the set {xyx,yUI} does not cover all of (0,1), hence for suitable c(0,1), the open set UcU 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 aR.
However, if I=(a,b) is bounded and UI=, 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 UR. Wlog. 0U. 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 RU is infinite.

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

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