Let $Z$ be the the set of dyadic and ternary rationals in the interval $\left[\frac12,1\right)$ whose 3-adic valuation is either $-1$ or $0$, with the standard absolute value topology inherited from the real line.

Let $X=\{z\in Z:\nu_3(x)=0\}$.

Let $Y=\{z\in Z:\nu_3(x)=-1\}$.

Now define an equivalence relation $\sim$ which partitions $Z$ into equivalence classes $\{x, y\}$ of cardinality $2$ with $x \in X$, $y \in Y$, and $f(x) = y$, where $f:X\to Y$ is given by $$f(x)=\begin{cases}\frac{4x}3 &\text{if}& x<\frac34\\ \frac{2x}3& \text{if}& x>\frac34.\end{cases}$$

So for example $\frac5{8}\sim\frac56$ and $\frac{7}{8}\sim\frac{7}{12}$.

The quotient (pseudo)metric $d_\sim$ on $Z/{\sim}$ is the infimum distance by which one can traverse from any equivalence class to another, by stepping on up to infinitely many equivalence classes in-between and summing only the distance between classes and not the distance travelled within classes. More formally this is defined at @EricWofsey's answer to "Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?".


I seek an explicit definition of $d_\sim$ in this instance. Of course, in a sense Eric's answer does give that, but it leaves me with a requirement to somehow iterate over all possible sequences of equivalence classes and determine the shortest path, something well beyond my capabilities.

Also Note

While this question stands alone without reference to the Collatz conjecture, I feel more comfortable declaring that identifying this metric is a component of my study of the conjecture, partly in the spirit of full-disclosure, but also becase it may be material to the answer, to be mindful of the following observations:

  • $g(x)=x+\frac132^{\nu_2(x)}$ is both a surjection $X\to Y$ and a surjection $Z/{\sim}\to Z/{\sim}$ and seen as a map $g:Z/{\sim}\to Z/{\sim}$ it is essentially the Collatz graph and its graph is connected if and only if the Collatz conjecture is true. One should not be surprised therefore, if some proof that $d_\sim$ is the trivial metric, $\forall [z_0],[z_1]:d_\sim([z_0],[z_1])=0$, were related to the claim that the graph of the orbit of $g$ through $Z/{\sim}$ is connected.

  • The $n$-indexed sequences of the form $s_n(x)=x+(1-2^{-6n})\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$ form an exact cover of $X$ (up to subsequences) and $g$ is their infinite limit. Moreover, for every $y\in Y$ there are precisely two $s_n$ (up to subsequences) whose union is the level set by $g$ of $y$.

  • If $d_\sim$ is not the trivial (pseudo)metric, it seems likely a proof that some sequence gives the infimum for the metric may use the sequences given in the two bullets above.

  • This question has a "dual", if you like, which is to enquire into the quotient pseudometric obtained when setting $x\sim' g(x)$ instead of $x\sim f(x)$ and moreover the conjecture is equivalent to the claim that $Z/({\sim}\cdot{\sim'})$ is a singleton having $Z$ as its only element.


1 Answer 1


One gets the trivial semi-distance. Let $d$ denote the semi-distance on $Z$ that defines $d_\sim$ on $Z/\sim$, that is $d_\sim([z],[z']):=d(z,z')$ for all $z$ and $z'$ in $Z$. Thus $d(z,z')\le |z-z'|$ for all $z$ and $z'$ in $Z$ and $d(x,f(x))=0$ for all $x\in X$.

It is convenient to extend the definition of $f$ to a self-map $[1/2,1)\to[1/2,1)$, still denoted $f$: $$f(x)=\begin{cases}\frac{4x}3 &\text{if}& 0\le x<\frac34\\ \frac{2x}3& \text{if}& \frac34\le x <1.\end{cases}$$

The key point is that all orbits of $f$ are dense. So for every $z$ and $z'$ in $Z$, and for every $\epsilon>0$, there is $m\in\mathbb N$ such that $\lvert f^m(z)-z'\rvert< \epsilon$. Since $X$ is also dense, and since $f$ is everywhere (right) continuous, for all indices $ 0\le j\le m$ there are $x_j\in X$ such that \begin{align*} \bigl\lvert f^j(z)-x_j\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m) \\ \bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m-1). \end{align*} Therefore for $0\le j\le m-1$ $$\bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \le \bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert+\bigl\lvert f^{j+1}(z)-x_{j+1}\bigr\rvert\le \frac{2\epsilon}m$$ and
$$d(z,z')\le\bigl\lvert z-x_0\bigr\rvert+ \bigg(\sum_{j=0}^{m-1} \bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \bigg) +\big|x_m-z'\big|\le 4\epsilon,$$ which proves that $d$ vanishes identically.

All orbits of $f$ are dense: Indeed, having all orbits dense is a property invariant by conjugation, and $f$ is conjugate to the irrational translation (modulo $1$) on $[0,1)\sim\mathbb{R/Z }$ given by $T_c:x\mapsto {x -c}$, where $c:=\frac {\log 3}{\log 2}-1$ (all orbits of a translation $T_c$ are dense iff $c$ is irrational).

Consider the homeomorphism $h:[0,1)\to[1/2,1)$ defined by $h(x)=2^{x-1}$. Then it is easy to check that $h\circ T_c=f \circ h$.

  • 1
    $\begingroup$ I think so (quotient being Hausdorff) but I do not see immediately. $\endgroup$ Jan 4, 2021 at 23:02
  • 1
    $\begingroup$ On the PS: To me, it was not obvious, and I wasn't even sure that the distance was trivial. How I proceeded, it's nothing special, but here it is. 1) It seemed that $X$ and $Y$ do not play a special role in studying these objects, and that it is healthier to have $f$ as a self-map, to make iterations. 2) one works better on the interval, than on the quotient. $\endgroup$ Jan 4, 2021 at 23:22
  • 1
    $\begingroup$ 3) in this semi-distance, $1/2$ is close to the right-end, for $f(3/4-\epsilon)= 1-4\epsilon/3$ and $f(3/4+\epsilon)= 1/2-2\epsilon/3$. So nothing changes if one starts from a circle, identifying the endpoints of $[1/2,1]$, instead using $[1/2,1)$, which makes $f$ a homeomorphism. This allows to use the tools of homeomorphims of the circle. I wrote $f$ as a time-1 flow to compute its rotation number and find a conjugation with a rotation: only at the end I realized the conjugation was so easy. $\endgroup$ Jan 4, 2021 at 23:28
  • 1
    $\begingroup$ PS: The quotient (semi)distance $\tilde d$ on $\tilde X$ for a quotient $\pi:X\to\tilde X$ is the maximum (semi)distance on $\tilde X$ that makes $\pi$ a $1$-Lipschitz map. This leads to the construction of $\tilde d$ via chains of jumps from a class to anothe (which I would call the construction rather than the definition of $\tilde d$ ). $\endgroup$ Jan 5, 2021 at 10:30
  • 1
    $\begingroup$ Yes, it was meant to be a proof (sorry if it wasn't clear). Having every orbit dense is a conjugation invariant (recall that $f=hgh^{-1}\Rightarrow f^m=hg^mh^{-1}$ for all $m$), and the translation $x\mapsto x+c\mod 1$ on $\mathbb{ T:=R/Z}$ does have every orbit dense if (and only if) $c$ is irrational (for the orbit of $x$ is $x+c \mathbb Z\mod 1 $), so every orbit of $f$ is dense too. $\endgroup$ Jan 5, 2021 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.