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.


Do these quotient spaces have a nontrivial topology?


$\Bbb R$ 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?


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.

  • $\begingroup$ 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? $\endgroup$ Dec 24 '20 at 8:54
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    $\begingroup$ Sorry, I did not realise you meant to include products. $\endgroup$ Dec 24 '20 at 8:58
  • $\begingroup$ Please don't delete. As the case of ternary rationals can still serve as a useful example for me to think about, which I plan to do today... if not correctable I may also be able to edit the question to suit your answer. $\endgroup$ Dec 24 '20 at 8:59
  • $\begingroup$ 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. $\endgroup$ Dec 24 '20 at 11:51
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    $\begingroup$ Apologies for the delay, I was unable to give some time to this for the past few days. Are we in agreement that $d([26],[10])=\left\lvert\dfrac{26}9-\dfrac{30}9\right\rvert=\dfrac49<\left\lvert\dfrac{26}9-\dfrac{10}9\right\rvert$ is a counterexample to your claimed proof? $\endgroup$ Jan 1 '21 at 14:47

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