Polynomials Are Continuous I R in Topology

A Topology such that the continuous functions are exactly the polynomials

Solution 1

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.

Solution 2

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.

Solution 3

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.

Comments

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.
- 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.
- 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.)
- @JimBelk, could you explain that process a bit more? That sounds intriguing.
- @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.
- (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.
- @JimBelk, that definition of $\psi$ (as a constant function) looks pretty funky. Is that what you intended?
- 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).
- Also, @JimBelk, what's wrong with having a topology finer than the usual one? The same topology is used for the domain and codomain.
- @dfeuer The definition of $\psi_r$ should be $\psi_r(x)=r(x-b)+b$. Also, the problem with a topology finer than the standard one was explained in my second comment: if closed rays really are closed, then you can paste together two polynomials to get a continuous function that isn't a polynomial.
- Am I correct in thinking that the Knaster-Tarski lemma proves that any topology on the field has a coarsest refinement on which polynomials are continuous?
- @dfeuer Could you be a bit more explicit?
- Dominik, I've put up an outline of my thoughts at proofwiki.org/wiki/User:Dfeuer/topext I do not claim to be certain that I am correct. I also do not claim that this will help resolve the question, but maybe something like it will.
- As it turns out the "lattice of topologies" is a complete (and even complemented) lattice, so that does work out, for whatever it's worth. One other thought: has anyone thought about whether there is a pair of topologies over the field that give just polynomials as continuous functions?
- Here's a specific suggestion. Let $U = \mathbb{R}-\mathbb{Z}$, and let $\mathcal{S} = \{p^{-1}(U) \mid p\in\mathbb{R}[x]\}$. Then $\mathcal{S}$ is a subbasis for a topology on $\mathbb{R}$. Does this topology have the requisite property?
- Here's another specific possibility: Let $p(y,f) = \{y,f(y),f(f(y)),\ldots\}$ for $y\in F$ and $f\in \Bbb R[x]$. Does that make a good set of closed sets, or can it be expanded to form one?
- @BenPasser, I don't see any particular reason to believe that we're looking for a topological field or even a topological ring, but maybe there is some way to show that this problem reduces to that one.
- @dfeuer I guess I can at least confirm that your thoughts on proofwiki are all sound, that is $\mathscr S$ really is (quite obviously in my eyes) a complete lattice, any fixed point of $h$ really is a topology making the given functions continuous, any such topology really is a fixpoint of both $f$ and $g$, hence of $h$, so, yes, the least fixed point is the coarsest topology such that the given functions are continuous and the given sets are open.
- @HagenvonEitzen: the weird thing about the situation (weird to me) is that either too fine or too coarse a topology will give too many continuous functions, so we're attempting a weird sort of balancing act. I read somewhere (I don't remember where just now) that there are examples of topologies under which the only continuous functions are the constant functions and the identity, so it's possible (at least for some sets) to get too picky.
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 $0$.
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.
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).
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.
@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.
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?
@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).
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).
What's that about "in the absence of a locally finite closed cover"? It sounds like you mean such a cover doesn't exist?
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?
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.
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.

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Source: https://9to5science.com/a-topology-such-that-the-continuous-functions-are-exactly-the-polynomials

Polynomials Are Continuous I R in Topology

A Topology such that the continuous functions are exactly the polynomials

Solution 1

Solution 2

Solution 3

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