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(I originally asked this question on Math.SE, where it received a lot of attention, but no solution.)

Which fields $K$ can be equipped with a topology such that a function $f:K \to K$ is continuous if and only if it is a polynomial function $f(x)=a_nx^n+\cdots+a_0$? Obviously, the finite fields with the discrete topology have this property, since every function $f:\Bbb F_q \to \Bbb F_q$ can be written as a polynomial. So what is with infinite fields?

I don't expect a full answer to this question, but I would even be satisfied if someone could show the existence or nonexistence of such a topology even for a single field such as $\Bbb Q, \Bbb R,\Bbb C,\Bbb Q_p$ or $ \Bbb F_q^\text{alg}$.

(Note that $(K,\tau)$ is not necessarily a topological field!) $$ $$

A short summary of the comments on Math.SE: Assume that you are given such a field $K$ with a topology $\tau$. Then $\tau$ is necessarily a $T_0$-space and connected. Also, the linear transformations $x \mapsto ax+b$ with $a \in K^\times$, $b \in K$ are homeomorphisms of $\tau$ (and there can be no other homeomorphisms). In the special case $K=\Bbb R$, for every $U \in \tau$ and $a \in \Bbb R$, we have $|(-\infty,a)\cap U |=|(a,\infty)\cap U |=2^{\aleph_0}$. More over, if there is an open interval $(a,b)$ and a $U \in \tau$ with $|(a,b)\cap U |<2^{\aleph_0}$, then for all $c,d \in \Bbb R$, $(-\infty,c)\cup (d,\infty) \in \tau$. As a proposal for a topology on $\Bbb R$, the coarsest topology such that all sets $p^{-1}(\Bbb Z)$ with $p \in \Bbb R[X]$ are closed, has been considered. Obviously every polynomial is continuous with respect to this topology. However, no one has given any reason why every continuous function in this topology should be a polynomial.

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    $\begingroup$ Dominik, have you read this meta thread? $\endgroup$
    – Asaf Karagila
    Jul 15, 2013 at 12:07
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    $\begingroup$ A necessary property that does not seem to be mentioned in the MSE thread is that for every finite $X\subseteq K$, $K\smallsetminus X$ is connected: if it were the union of disjoint nonempty (and therefore infinite) open sets $U,V$, the function $$f(x)=\begin{cases}0&x\notin U\\\prod_{a\in X}(x-a)&x\notin V\end{cases}$$ would be continuous, but not a polynomial. $\endgroup$ Jul 15, 2013 at 13:31
  • $\begingroup$ @Emil: Doesn't that imply that the topology has to be a subset of the cofinite topology, then? $\endgroup$
    – Asaf Karagila
    Jul 15, 2013 at 14:01
  • $\begingroup$ @Asaf: I don’t see why, the usual topology on $\mathbb R^n$ also has this property for $n\ge2$, as well as plenty other spaces. The topology is a superset of the cofinite topology, as it is $T_1$. $\endgroup$ Jul 15, 2013 at 14:43
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    $\begingroup$ For $\mathbb{C}$, there is an easy solution if you only consider entire function instead of continuous ones. The only entire functions with a removable singularity at $\infty$ are indeed the polynomials. $\endgroup$
    – Marc Palm
    Jul 16, 2013 at 12:18

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In the case of a topological field (edit: containing real numbers) the answer is not. Let $K$ a infinite field, with a such topology $\tau$ coherent with the algebraic field structure then $(K, +, 0, \tau)$ is a topological groups. As you said $G$ must be $T_1$ (i.e. $\{0\}$ is closed, this is equivalent to $T_0$ in a topological group), then it is completely regular, i.e. continuous functions separate closed set from points, but if $K$ is infinite the topology $\tau$ cannot be discrete or indiscrete (in these cases any function become continuous, but a polynomial can have only a finite set of zeros , then exists a lot of non polynomial functions), neither $\tau$ could be cofinite (i.e. opens are exactly complements of finite sets) because cofinite topology in a infinite set isn't $T_2$. Then exist a infinite closed $A$ and a $x\not\in A$ and a continuous function that separate these, but just because $A$ is infinite this cannot be polynomial.

I wanted to write this answer as a comment, but it was too long

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    $\begingroup$ OP pointed out that the topology need not turn $K$ into a topological field. $\endgroup$
    – Rasmus
    Jul 15, 2013 at 11:59
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    $\begingroup$ I don't understand. Complete regularity has to do with separation by continuous real-valued functions. How does that relate to continuity of $K$-valued functions? $\endgroup$ Jul 15, 2013 at 19:24
  • $\begingroup$ Diepeveen, you right, my answere (would be only a comments) nedd some precisations (edits). $\endgroup$ Jul 16, 2013 at 10:19
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    $\begingroup$ "containing real numbers" is still very imprecise. What you need to make your argument work is a continuous injection of the real numbers with the order topology into $(K, \tau)$. Even in the case $K = \mathbb{R}$ you need to assume that $\tau$ is coarser than the order topology before you can conclude that it is not completely regular. So far I have only seen proof that it cannot be finer, not that it must be coarser. $\endgroup$ Jul 16, 2013 at 21:12
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    $\begingroup$ Another weird topology is obtained by taking an automorphism $\sigma$ of $\mathbb C$ other than identity or complex conjugation, and transfer to $\mathbb R$ the standard topology from $\sigma(\mathbb R)$. $\endgroup$ Jul 17, 2013 at 11:43

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