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Let $T: X \to X$ be a continuous map on a compact metric space $X$. We say $T$ is distal if $\inf_n d(T^n x, T^n y) = 0$ implies $x = y$.

Then it is true that $T$ is bijective.

Question: Is there an elementary proof of this fact? (Injectivity clearly follows, surjectivity is the issue.) The two proofs I know go through the enveloping semigroup, and the Stone-Čech compactification $\beta \mathbb N$ respectively.

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    $\begingroup$ That's a very nice result. Do you have a reference for the Stone Cech compactification proof? $\endgroup$
    – Asvin
    May 25, 2021 at 1:38
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    $\begingroup$ It is! You can find the proof in Bergelson’s survey article Ergodic Ramsey Theory - an Update. It’s on page 33. $\endgroup$
    – Nate River
    May 25, 2021 at 1:45
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    $\begingroup$ Wow! Very surprising. I would expect that this is a very elementary exercise. $\endgroup$ May 25, 2021 at 2:50
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    $\begingroup$ What is $X$ is say a circle or a Euclidean sphere? Is it easier then? It would be much more surprising if the result was equally difficult for a simple space. $\endgroup$ May 25, 2021 at 15:00
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    $\begingroup$ @YCor, I think the definition of distal for topological semigroups is slightly more complicated. You can find it in arxiv.org/pdf/1708.00996.pdf and the conclusion is you have a group. They show the whole enveloping semigroup is a group as well $\endgroup$ May 25, 2021 at 16:26

3 Answers 3

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The answer is yesser than I thought. I mentioned this issue at http://eventos.cmm.uchile.cl/edynamicsxiii/, since the proximality lemma from my previous answer was discussed there. Someone pointed out that Hindman's original proof of his famous theorem is at least somewhat elementary in some technical sense, and implies a significant part of the proximality theorem, so the answer must be "yes" at least in some technical sense.

After a bit of searching I found the relevant paper [1]: in the sense of reverse mathematics, your theorem is provable in $\mathrm{ACA}_0^+$, a certain fragment of second-order arithmetic.

I quote the relevant theorem in the form stated in this paper. They call this the Auslander-Ellis theorem.

Theorem. Let $X$ be a compact metric space and let $T : X \to X$ be continuous. Regard $(X,T)$ as dynamical system. Given $x \in X$, there exists $y \in X$ such that $y$ is uniformly recurrent and proximal to $x$.

They show that this theorem is provable in $\mathrm{ACA}_0^+$. Let me recall how to conclude your result (this deduction seems very elementary, so I guess it needs much less than $\mathrm{ACA}_0^+$).

Corollary. Every distal system is invertible.

Proof. Injectivity is clear. Take $x \in X$, and apply the previous theorem, to get that $x$ is proximal to some uniformly recurrent $y$. Then $x = y$, so every point in $X$ is uniformly recurrent. Clearly this implies surjectivity, since if $U$ is a neighborhood of any $x \in X$, we have $T^n(x) \in U$ for some positive $n$, and then $T^{n-1}(x)$ maps to $U$ in $T$, so $TX$ is dense in $X$ and by compactness is equal to $X$. Square.

I am not an expert, but as far as I understand, second-order arithmetic in itself is weaker than Zermelo-Frankel set theory (without choice), and $\mathrm{ACA}_0^+$ is about medium strength as far as the most commonly studied fragments of second-order arithmetic go. Roughly, the logic $\mathrm{ACA}_0$ says that a set of natural numbers exists if you can define it by an arithmetical formula, and $\mathrm{ACA}_0^+$ adds the $\omega$th Turing jump of each set already definable.

Nevertheless, to quote [1], "It is well known that all existing proofs of HT are nonconstructive. One of the goals of this paper is to delimit the degree of nonconstructivity which is inherent in Hindman's Theorem."

To summarize the logical connections obtained: Write AET for the theorem about proximality, HT for Hindman's theorem, DT for OP's theorem about distal systems. We have $$\mathrm{ACA_0^+} \implies \mathrm{AET} \implies \mathrm{HT} \implies \mathrm{ACA_0} \wedge \mathrm{DT}$$ Intuitively (and again I am not an expert), if you believe results of a certain flavor of infinite computation are well-defined, you can prove AET and therefore the theorem about distal systems. But in theory, the problem about distal systems could be much easier (I have a hard time imagining a proof not going through proximal pairs, but I've been wrong before). In the reverse math framework, one could ask if it is in $\mathrm{ACA}_0$, or $\mathrm{WKL}_0$, or $\mathrm{RCA}_0$, in decreasing order of strength.

Reference

[1]: Blass, Andreas R.; Hirst, Jeffry L.; Simpson, Stephen G., Logical analysis of some theorems of combinatorics and topological dynamics, Logic and combinatorics, Proc. AMS-IMS-SIAM Conf., Arcata/Calif. 1985, Contemp. Math. 65, 125-156 (1987). ZBL0652.03040.

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  • $\begingroup$ Wow. So the logic stuff is a little bit beyond me, but I guess it means something like the proofs are not “that” nonconstructive? Maybe it implies something about how large the spaces used in the proof have to be. Though I’m way out of my league here so I’ll just keep wondering for now. $\endgroup$
    – Nate River
    Jun 16, 2021 at 10:42
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    $\begingroup$ Second-order arithmetic means that you can only talk about sets of natural numbers in the proof. For the proximality theorem, the proof through Ellis semigroup and the proof in Auslander's book talk about subsets of some uncountable set, so they are not directly formalizable in this language. Of course you'll have to encode your system into sets of numbers to even be able to talk about the statement of the theorem, and sometimes there are encoding issues, but I guess not in this case since it was not discussed in the introduction of the paper. $\endgroup$
    – Ville Salo
    Jun 16, 2021 at 11:59
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It is a very nice result ! Let me try an "elementary" answer.

Define $\delta: X\times X \to \mathbb R_+$ by $$\delta(x,y)= \inf_{n\in \mathbb N} d(T^n(x), T^n(y))~.$$ This function is continuous and vanishes exactly on the diagonal by distality.

Edit: This function is NOT continuous in general, as pointed out in the comments.

Set $$F= \bigcap_{n\in \mathbb N} T^n(X)~.$$ $F$ is compact and for all $x\in X$ we have $d(T^n(x), F) \underset{n\to +\infty} \longrightarrow 0$. Since $$\inf_{y\in F} \delta(x,y) \leq \inf_{y\in F} d(T^n(x),y)$$ for all $n$, we conclude that $$\inf_{y\in F} \delta(x,y) = 0~.$$

By compactness of $F$ and distality, we conclude that $x\in F$ for all $x$. Hence $F=X$ and $T$ is surjective.

Edit: Of course this last line falls appart if $\delta$ is not continuous. So the question is: why is the infimum $\inf_{y\in F} \delta(x,y) = 0$ is attained ?

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    $\begingroup$ Hmm, maybe I’m missing something - is it obvious that $\delta$ is continuous? Naively this is a pointwise infimum of continuous functions which need not be continuous.. $\endgroup$
    – Nate River
    May 25, 2021 at 10:56
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    $\begingroup$ The Cantor set $2^{\mathbb N}$ with its usual metric saying two infinite words are close if they have a long common prefix and the shift map $T$ (eliminating the first letter of an infinite word) is a counterexample to $\delta$ being continuous. let $z$ be a fixed infinite word over the alphabet $\{0,1\}$ and consider the sequences of infinite words $u_n=0^nz$ and $v_n=1^nz$. Then $u_n$ converges to $0^{\infty}$ and $v_n$ converges to $1^{\infty}$. Clearly $\delta(u_n,v_n)=0$ all $n$ since $T^n(u_n)=T^n(v_n)$ but $\delta(0^\infty,1^\infty)\neq 0$. $\endgroup$ May 25, 2021 at 12:09
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    $\begingroup$ Right. Well, actually I suppose the most basic example of rotating points on a disk by their distance from the center is a counterexample. Using the unit disk of $\mathbb{C}$ as the model, $\delta(x, y) = 2$ for $x,y$ on the boundary, but nearby it can be arbitrarily small, just take two points on different circles. $\endgroup$
    – Ville Salo
    May 25, 2021 at 12:21
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    $\begingroup$ As pointed out in the comments, I was too quick to assert that $\delta$ is continuous. So there is really nothing to save... $\endgroup$ May 25, 2021 at 17:45
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    $\begingroup$ Recently I took an introductory course on topological dynamics, and this was (mistakenly) one of the exercises there. The intended "proof" was precisely this, so at least you're not the first person to make this error :) $\endgroup$
    – Random
    May 25, 2021 at 19:24
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edit

My appeal to authority in the following answer makes no sense, because it seems Auslander's flows are actually assumed to be invertible. My own intuition is still that the issue is the same, so I'll keep this for now.

original

I also did not understand why $\delta$ is continuous in the answer of Nicolas Tholozan, maybe they will clarify and I'll delete this, but in the meantime, I would have guessed this has no known elementary proof (or at least didn't in '88...) for the following reason: The only proof I can see is through the following theorem on page 67 in Auslander's '88 book Minimal Flows and their Extensions (in the chapter on distal flows):

Let $(X, T)$ be a flow and let $x \in X$. Then there is an almost periodic point $x^*$ which is proximal to $x$.

Your result easily follows, because the almost periodic point will be in the eventual image. Auslander makes the following comment after the theorem:

"Every known proof of this theorem requires the use of a "large" product space (another proof will be given in the next chapter). It would be interesting to find a direct proof."

I don't see a reduction of this to the result you are after, so possibly yours is easier, but this was the only reasonable approach I could see. I also don't know if Auslander's comment has already been addressed somewhere, I am not an expert on distality by any means.

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    $\begingroup$ This recent paper by Auslander arxiv.org/pdf/1708.00996.pdf is still using Ellis semigroups for the more general case of distal flows of semigroups but I didn’t see any statement that other methods are known. $\endgroup$ May 25, 2021 at 12:14

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