# Does the quotient of the dyadic and ternary rationals greater than $1$ by their powers of $3$ have the trivial topology?

Consider two topological spaces:

1. Begin with $$X_{>0}$$ the positive dyadic and ternary rationals (including products e.g. $$\frac n6$$), with their topology as a subspace of $$\Bbb R$$. Then consider the quotient space $$X_{>0}/\langle3\rangle$$ given by completing the equivalence relation $$x\sim 3x$$ or in other words:

$$x\sim y\iff\exists i\in\Bbb Z:3^ix=y$$

2. As above, but now begin with $$X_{>1}$$ the dyadic and ternary rationals (including products) greater than $$1$$ and take the quotient $$X_{>1}/\langle3\rangle$$ using the same equivalence relation.

Question

Do these quotient spaces have a nontrivial topology?

Attempt

$$\Bbb R$$ is a metric space. Therefore the quotient topology is the same as the toplogy given by the quotient pseudometric, right? Or are there other Hausdorff quotient topologies?

Then for case 1: its topology IS trivial because I can pick arbitrarily large $$i$$ such that $$d(3^{-i}x,3^{-i}y)$$ is arbitrarily small in the vicinity of $$0$$.

For case 2: There's a prima facie argument that this quotient is Hausdorff because for any pair $$x,y$$ there's some maximal $$i_1,i_2$$ such that $$\lvert 3^{i_1}x-3^{i_2}y\rvert$$ is a difference between two representatives greater than $$1$$ - guaranteeing disjoint vicinities around them. But this isn't conclusive because correctly using the quotient metric involves finding the shortest distance between two elements in $$X_{>1}/\langle3\rangle$$ using any sequence of stepping stones in-between. I'm pretty much baffled how to do that. How do I determine a non-recursive function for the pseudometric?

Update

For the reasons given in this answer: https://mathoverflow.net/a/380398/91341 which is largely the same as this problem, I am satisfied that the quotient pseudometric is the trivial $$\forall x,y:d(x,y)=0$$. I am of the view however that the quotient topology itself is NOT trivial, although I cannot show it.

Let $$[x],[y]\in X_{>1}/\langle3\rangle$$ with $$x,y\in X_{>1}$$ the least element of their respective equivalence classes. I claim that $$\begin{equation} \label{eq:pseudometric} d([x],[y])=|3^{-i}x-3^{-j}y| \tag{1} \end{equation}$$ where $$i$$ and $$j$$ are the unique nonnegative intergers for which $$1\leq 3^{-i}x,3^{-j}y<3$$. Note that $$3^{-i}x$$ and $$3^{-j}y$$ are only contained in $$[x]$$ and $$[y]$$ if they are ternary. However, since the ternary rationals are dense in $$\mathbb{R}$$ it suffices to prove (1) for both $$x$$ and $$y$$ ternary. Accordingly, suppose both $$x$$ and $$y$$ are both ternary and the least element of their equivalence class. Thus, $$1\leq x, y<3$$ so that (1) asserts that $$d([x],[y])=|x-y|$$. If $$a\sim x$$ and $$b\sim y$$ with $$a\neq x$$ then $$|a-b|>|x-y|$$ so that $$|x-y|<|x-a|+|a-b|+|b-y|$$. Thus, the infimum of all the lengths of chains from $$x$$ to $$y$$ is $$|x-y|$$ so that (1) holds.
• Apologies, I realise I have been ambiguous at best and misleading at worst in my question. Did you include e.g. $\frac76$ in $X$ which is a product of both dyadic and ternary rationals? Dec 24, 2020 at 8:54
• For each $x\in X_{>1}$ there exists a unique nonnegative integer $i$ for which $1\leq 3^{-i}x<3$. Note that $x^*=3^{-i}x\in [x]$ and is the least element. Let $[x],[y]\in X_{>1}/\langle3\rangle$. I claim that $$d([x],[y])=|x^*-y^*|. \tag{1}$$ Let $\{a_0,a_1,\dots,a_n\}$ and $\{b_0,b_1,\dots,b_n\}$ be any chain between $x$ and $y$. Then, $$\sum_{i=0}^{n-1} |b_i-a_{i+1}| \geq \sum_{i=0}^{n-1} |b_i^*-a_{i+1}^*| =|x^*-a_1^*|+|b_1^*-a_1^*|+\cdots +|b_{n-1}^*-y^*|\geq |x^*-y^*|.$$ Hence, (1) holds. Dec 24, 2020 at 11:51
• Apologies for the delay, I was unable to give some time to this for the past few days. Are we in agreement that $d(,)=\left\lvert\dfrac{26}9-\dfrac{30}9\right\rvert=\dfrac49<\left\lvert\dfrac{26}9-\dfrac{10}9\right\rvert$ is a counterexample to your claimed proof? Jan 1, 2021 at 14:47