I was wondering 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. 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 $\Bbb Q, \Bbb R,\Bbb C$ or $ \Bbb F_q^\text{alg} $). 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 $\tau$. Then for $a,b \in K$ , $a \ne 0$, $x \mapsto ax+b$ is a continuous function with continuous inverse, hence a homeomorphism. Thus $K$ is a homogenous space with doubly transitive homeomorphism group. Since $\tau$ cannot be indiscrete, there is an open set $U$, and $x,y \in K$ with $x \in U,y \not\in U$. Now for every $a \in K$, $a*(U-y)/x$ includes $a$ but not $0$, and thus $K\setminus\{0\}=\bigcup_{a \in K^\times}a*(U-y)/x$, 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 $f:K \to \{0,1\} \subset K$, 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.

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    $\begingroup$ If every polynomial function $f : K \to K$ is continuous and $\{ 0 \}$ is closed, then the topology on $K$ must be finer than the Zariski topology. However, the Zariski topology does not have the desired property: for example, the absolute value function $\mathbb{Q} \to \mathbb{Q}$ is Zariski-continuous. $\endgroup$
    – Zhen Lin
    Commented Jun 23, 2013 at 19:08
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    $\begingroup$ As proved by rohit in the comments to this question, any nonempty open subset of $\mathbb{Q}$ or $\mathbb{R}$ under the desired topology will have to be unbounded. (If there's a bounded open set, you can translate it around using affine maps to get it to fill up an open interval, which gives you that every open interval is open, which leads to a contradiction since a piecewise-polynomial function isn't polynomial.) $\endgroup$
    – Jim Belk
    Commented Jun 23, 2013 at 20:51
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    $\begingroup$ @dfeuer Suppose $U$ is a bounded open set, and let $a,b\in\mathbb{R}$ so that $a<x<b$ for all $x\in U$. For each $r\in(0,1)$, let $\phi_r(x) = r(x-a)+a$, and let $\psi_r(x) = (x-b)+b$. Then each $\phi_r$ and $\psi_r$ are homeomorphisms (being polynomials whose inverses are also polynomials), and $\left(\bigcup_{r\in(0,1)} \phi_r(U)\right)\cup\left(\bigcup_{r\in(0,1)} \psi_r(U)\right) = (a,b)$, so the open interval $(a,b)$ is an open set. $\endgroup$
    – Jim Belk
    Commented Jun 25, 2013 at 1:47
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    $\begingroup$ (Continued) It follows that any open interval $(c,d)$ is an open set, since $(c,d)$ is the image of $(a,b)$ under an affine linear map, and all affine linear maps are homeomorphisms. It follows that every subset of $\mathbb{R}$ that is open under the usual topology is open under the new topology, and similarly for closed sets. Then it follows from the pasting lemma that piecewise functions like $f(x) = \begin{cases}x & \text{if }x \leq 0 \\ x^2 & \text{if }x\geq 0\end{cases}$ must be continuous, a contradiction. $\endgroup$
    – Jim Belk
    Commented Jun 25, 2013 at 1:52
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    $\begingroup$ Are there any results on the cardinality of $C(K, K)$ or even $Homeo(K, K)$ for topological fields or groups? I ask because the cardinality of the set of polynomial functions for an infinite field $K$ is $|K|$ (since $|Polyn(K)| \leq \sum |K^n| = \sum |K| = |K|$), but I would expect there to be more continuous functions than that. Perhaps the cardinality of $C(K, K)$ can be bounded below due to the complete regularity of $K$ (just a wild guess). $\endgroup$
    – Ben Passer
    Commented Jun 27, 2013 at 18:12

5 Answers 5


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

Consider the case $K=\mathbb R$. Such a topology $\mathcal T$ has to be invariant under translations, scaling, and reflection because $x\mapsto ax+b$ with $a\ne 0$ is a homeomorphism.

(cf. JimBelks comments above) Assume there is a nonempty open set $U$ bounded from below, then wlog. (by translation invariance) $U\subseteq (0,\infty)$ and hence $(0,\infty)=\bigcup_{r>0}rU$ is open. By reflection, also $(-\infty,0)$ is open and by the pasting lemma, $x\mapsto|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x\le 0\end{cases}$ 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 form$I=(-\infty,a)$, $I=(a,\infty)$, or $I=(a,b)$. Assume $|U\cap I|<|\mathbb R|$ and $0\notin I$. Then the set $\{\frac xy\mid x,y\in U\cap I\}$ does not cover all of $(0,1)$, hence for suitable $c\in(0,1)$, the open set $U\cap cU$ is disjoint from $I$ and nonempty (contains $0$). If $I$ itself is unbounded this contradicts the result above. Therefore (by symmetry) $$|U\cap(a,\infty)|=|U\cap(-\infty,a)|=|\mathbb R|$$ for all nonempty open $U$ and $a\in\mathbb R$. However, if $I=(a,b)$ is bounded and $U\cap I=\emptyset$, then $\bigcup_{ca+d<a\atop cb+d>b} (cU+d) =(-\infty,a)\cup (b,\infty)$ is open. Taking inverse images under a suitable cubic, one sees that all sets of the form $(-\infty,a)\cup c,d)\cup (b,\infty)$ are open and ultimately all open neighbourhoods (in the standard topology!) of the point at infinity in the one-point compactification of $\mathbb R$ 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:

$|U\cap (a,b)|=|\mathbb R|$ for all nonempty $U$ and bounded intervals $(a,b)$, or all (standard) open neighbourhoods of $\infty$ are open.

Since $x\mapsto |x|$ is continuous under the indiscrete topology, there exists an open set $\emptyset\ne U\ne \mathbb R$. Wlog. $0\notin U$. Then $\bigcup_{c>0}cU=K\setminus\{0\}$ is open, hence

points are closed.

So $\mathcal T$ is coarser than the cofinite topology. Since $x\mapsto |x|$ is continuous under the cofinite topology, it is strictly coarser, i.e. there exists an open set $U\ne \emptyset $ such that $\mathbb R\setminus U$ is infinite.

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    $\begingroup$ Should this be a CW answer to allow others to work on simplifying/clarifying/expanding? We've already discussed that points must be closed and that the space must be connected. It was just a mind-glitch on my part that made me fail to realize that no bounded open sets means no bounded-above/below open sets. Also: what can be done with the fact that every polynomial has a finite number of zeroes? $\endgroup$
    – dfeuer
    Commented Jul 9, 2013 at 22:05
  • $\begingroup$ @dfeuer Yes, I was hoping for more results while writing down this answer, but all hopes faded away while I continued writing. Meanwhile I think that JimBelk's idea with the sets $p^{-1}(\mathbb R\setminus \mathbb Z)$ as subbasis will either work or show that it is not possible; at least $\mathbb Z$ is the simplest idea of an infinite closed set as mentioned in my last sentence and it is about getting time to make use of all polynomials, not just linear (well, and cubic in one step). (I think, your Knaster-Tarski idea is not necessary here because polynomials are closed under composition). $\endgroup$ Commented Jul 10, 2013 at 7:52
  • $\begingroup$ The Knaster-Tarski idea is about getting a sense of the shape of the portion of the lattice of topologies that we are interested in. It doesn't necessarily do anything useful, but I'm thinking vaguely about how to hem in the potentially-useful part of that lattice. $T_1$ bounds it below. Connected restricts it above. K-T means that we can always move up just a bit from anywhere to make polynomials continuous (but that may not be useful). $\endgroup$
    – dfeuer
    Commented Jul 10, 2013 at 8:55

In a comment on the original post, Jim Belk proposed the topology on $\mathbb{R}$ with a subbasis given by the collection of all sets of the form $\mathbb{R}\setminus p^{-1}(\mathbb{Z})$, where $p\in\mathbb{R}[x]$.

Alas, this topology does not work. In particular, I claim that the function $f(x)=\arctan x$ is continuous w.r.t. this topology.

Proof: As pointed out in another answer here, a polynomial takes only finitely many integer values on any bounded set. So $p^{-1}(\mathbb{Z})\cap(-\pi/2,\pi/2)$ is finite for any polynomial $p$. It follows that for any proper closed set $C$, we have that $f^{-1}(C)$ is a finite set and therefore closed (because this topology is $T_1$). Q.E.D.

Note: We can avoid this difficulty by using, say, the rationals, or the dyadic rationals, in lieu of the integers.


Some hopefully-correct thoughts about Jim Belk's proposal for the real numbers:

Polynomials in that topology are continuous, because for polynomials $p$ and $q$, $q^{-1}(p^{-1}(\Bbb Z))=(p\circ q)^{-1}(\Bbb Z)$.

The topology is coarser than the usual one, so the space is connected. It's also $T_1$, because $\Bbb Z+x$ and $\pi \Bbb Z+x$ are closed for each $x$. So it at least meets the basic requirements we've laid out already.

Since the preimage of each integer under a given polynomial is finite, each sub-basic closed set is countable, so in fact every closed set is countable. We know then, that the topology must be somewhere between cofinite and cocountable. It is easy to see that it must be strictly between, because the nonconstant continuous functions in the cofinite (cocountable) topology are those which take no value more than finitely (countably) often.

Moreover, a polynomial takes only finitely many integer values on any bounded set, so in fact the subspace topology on any bounded set is cofinite. This also means that any nontrivial closed set is order-isomorphic to a subset of $\mathbb{Z}$.

We can also conclude, since any two nonempty subsets have nonempty intersection, that it is hyperconnected, therefore not $T_2$, and thus not a topological group under any operation. Furthermore, no finite collection of closed sets covers any nonempty open set, so in the absence of a locally finite closed cover, even the strongest form of the pasting lemma does not apply.

  • $\begingroup$ What's that about "in the absence of a locally finite closed cover"? It sounds like you mean such a cover doesn't exist? $\endgroup$
    – dfeuer
    Commented Jul 14, 2013 at 18:04
  • $\begingroup$ Also, where do we go from here? Can we check, for instance, whether all functions continuous under this topology are continuous under the usual one? What about our old favorite, absolute value? $\endgroup$
    – dfeuer
    Commented Jul 15, 2013 at 2:13
  • $\begingroup$ For a locally finite closed cover to exist, some neighbourhood of each point must be covered by finitely many closed sets. In this topology all neighbourhoods are uncountable, while all nontrivial closed sets are countable. $\endgroup$ Commented Jul 17, 2013 at 10:35
  • $\begingroup$ What would be invaluable at this point, is a simpler characterization of closed sets, especially an easier way to see that certain sets are not closed. For example, I can prove without a problem that the absolute value is discontinuous, assuming that $\mathbb{Z}_+$ is not closed. Unfortunately that assumption does not seem to be so easy to prove. $\endgroup$ Commented Jul 18, 2013 at 11:04

Below I am showing a property that will show how "big" open sets must be in this topology.

For $K=\mathbb{R}$. If $v\not=0$ and $U$ is an open neighbourhood of $v$, then there exists $v'\in U$, such that $vv'<0$.

Proof. Let $f(x)=|x|$. Since $x\mapsto -x$ is homeomorphism, for any open $V$, $-V$ is open. For any open neighbourhood $V$ of $0$, $0\in V\cap-V$ and $f(V\cap-V)\subset V$. Thus $f$ is continuous at $0$. But $f$ is discontinuous. Let $z$ be a point of discontinuity of $f$. We know that $z\not=0$. Take any open neighbourhood $G$ of $f(z)$. Notice that if all elements of $G$ has the same sign as $z$, then (for $z>0$:$f(G)=G$ and $z\in G$) or (for $z < 0$:$f(-G)=G$ and $z\in -G$). From that observation follows that if there exists an neighbourhood of $z$ whose elements have the same sign, then $f$ is continuous at $z$. So each open neighborhood of $z$ must contain an element that is of opposite sign. Now, take any $v\not=0$, let $\alpha=\frac{z}{v}$. Keep in mind that $x\mapsto \alpha x$ is a homeomorphism. Take any open neighbourhood $U$ of $v$. Note that $z\in \alpha U$. Thus there is $e\in \alpha U$ such that $ez<0$. But then $\frac{e}{\alpha}\in U$ and $\frac{ez}{\alpha^2} = \frac{e}{\alpha}v < 0$. So put $v'= \frac{e}{\alpha}$. This completes the proof.

And I think that because of $x\mapsto x + b$ and $x\mapsto -x$ are homeomorphisms we can easily extend this to the following:

For $K=\mathbb{R}$. For any $v$ and any $M>0$, each neighbourhood of $v$ contains point $v'$ for which $v'-v > M$.

  • $\begingroup$ Don't you mean that there exists an open neighborhood of $z$ containing an element with sign opposite that of $z$, rather than that all open neighborhoods of $z$ have such? Your conclusion that every open neighborhood is unbounded is the same as Jim Belk's conclusion. The part of your post that struck me as interesting is that the absolute value function has to be continuous at $0$. $\endgroup$
    – dfeuer
    Commented Jul 8, 2013 at 16:48
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    $\begingroup$ Wait, I see now that you're saying something a little different, that each neighborhood is unbounded above, but I still don't think you've quite shown that. $\endgroup$
    – dfeuer
    Commented Jul 8, 2013 at 16:56
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    $\begingroup$ no, this is for all neihgbourhoods of $z$, because if any of them will have all elements of the same sign, you can just prove continuity of $x\mapsto |x|$ at point $z$ (because it is the same as $x\mapsto x$ or $x\mapsto -x$ on such a neighbourhood). $\endgroup$
    – Michal
    Commented Jul 8, 2013 at 17:42
  • $\begingroup$ But there's no inherent problem with it being continuous at some point other than zero. We only need for there to be a point where it is not continuous. $\endgroup$
    – dfeuer
    Commented Jul 8, 2013 at 17:48
  • $\begingroup$ @dfeuer $U$ is unbounded above can also be shown with JimBelk's method: Assume $U\subseteq (-\infty,a)$. Then by affine linear homeomorhisms, all $c(U-a)+a$ with $c>0$ are open, hence their union $(-\infty,a)$ is open and by translation any $(-\infty,b)$ is open. Then also $-(-\infty,-a)=(a,\infty)$ is open and also $(a,b)=(a,\infty)\cap (-\infty,b)$. Anyway, the pasting lemma applies. $\endgroup$ Commented Jul 9, 2013 at 14:19

I thought I’d assemble some of the thoughts so far, plus add a few more.

Suppose that $F$ is a field, and that $\mathcal{T}$ is a topology on $F$ such that the set of continuous functions from $F$ to $F$ equals the set of polynomial functions with coefficients in $F$.

Then, as noted elsewhere on this page:

  • $F$ is not discrete.

  • $F$ is $T_1$.

  • Some infinite proper subset of $F$ is closed.

  • Every nonempty open set is infinite.

  • If $F=\mathbb{Q}$ or $F=\mathbb{R}$, then every nonempty open set is unbounded.

  • $F$ is bihomogeneous.

  • The homeomorphism group of $F$ is $\{ax+b\;|\;a,b\in F\}$.

  • $F$ is connected.


  • $F$ is crowded. In other words, $F$ has no isolated points. (Proof: Otherwise $F$ would be discrete.)

  • If $S$ is a finite subset of $F$, then $F\setminus S$ is connected. (Proof: If $S$ is empty, this follows because $F$ is connected. Otherwise, suppose $U\cup V$ is a disconnection of $F\setminus S$. Let $p(x)=(x-s_1)\cdots(x-s_n)$, where $S=\{s_1,\dots,s_n\}$. The function which equals $0$ on $U$ and on $S$ but equals $p$ on $V$ is continuous by the pasting lemma.)

  • If $F=\mathbb{R}$, then $F$ is not compact Hausdorff. (Proof #1: Suppose $F$ is compact Hausdorff. Then $[a,\infty)$ is closed, being the image of the compact set $F$ under the map $x^2+a$. The collection of these closed sets has the finite intersection property and therefore has nonempty intersection, which is a contradiction.) (Proof #2: The set $[0,\infty)$ being closed contradicts $(0,\infty)$ being dense---see the last bullet point, below.) (Proof #3: The function $f(x)=x^3$ is continuous and bijective, but not a homeomorphism, because its inverse function is not continuous.) Note that the logic of the first two proofs applies not just to $\mathbb{R}$ but to any ordered field in which every positive element is a square.

  • Regarding $F$ as a group under addition, the closure of any subgroup is a subgroup. (Idea of proof: Use that translations are homeomorphisms, as is the function $f(x)=-x$.) Many statements of this form hold, for example, replacing “closure” with interior or replacing “subgroup” with “set closed under multiplication,” etc.

  • Again regarding $F$ as a group under addition, every nonempty open set generates $F$. (Proof: Otherwise, let $U$ be an open set. The subgroup $H$ generated by $U$ is a union of open sets. If $H$ is a proper subgroup, then, considering the cosets of $H$, we see that this violates the connectivity of $H$.)

  • Every subgroup of $F$ is either dense or nowhere dense. (Proof: Follows from the last item.)

  • If $F$ is an ordered field, then every open set containing $0$ contains both positive and negative elements. (Proof: Let $U$ be an open set containing $0$. We know $0$ is not an isolated point, so $U$ contains some other point $a$. If $U$ contains only non-negative elements, then $[0,\infty)$ is the union of the sets $U+b$ where $b\geq 0$ and is therefore open. But then $(-\infty,0)$ is a union of the open sets $(-\infty,c]$. This contradicts $F$ being connected. Similarly if $U$ contains only non-positive elements.)

  • If $F$ is an ordered field, then $A=(0,\infty)$ is dense in $F$. (Proof: From the last item, we see that $0$ is in $\overline{A}$, the closure of $A$. But $[0,\infty)$ cannot be closed, else $F\setminus\{0\}$ is disconnected. So $\overline{A}$ contains a negative element $b$. But then $\overline{A}=F$, because $A$, and therefore $\overline{A}$, is closed under multiplication.)


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