What is the quotient of the absolute value metric in $\Bbb Z[\frac16]^+/\langle2,3\rangle$?

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What is the quotient of the absolute value metric in $\Bbb Z[\frac16]^+/\langle2,3\rangle$?

I'm somewhat baffled by Wikipedia's definition of a quotient pseudometric. How does it apply to the example given below of $\Bbb Z[\frac16]^+/\langle2,3\rangle,d'$?

From Wikipedia I have the definition:

If $M$ is a metric space with metric $d$, and $\sim$ is an equivalence relation on $M$, then we can endow the quotient set $M/{\sim}$ with the following (pseudo)metric. Given two equivalence classes $[x]$ and $[y]$, we define $$ d'([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} $$ where the infimum is taken over all finite sequences $(p_1, p_2, \dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $[p_1]=[x], [q_n]=[y], [q_i]=[p_{i+1}], i=1,2,\dots, n-1$. In general this will only define a pseudometric, i.e. $d'([x],[y])=0$ does not necessarily imply that $[x]=[y]$. However for nice equivalence relations (e.g., those given by gluing together polyhedra along faces), it is a metric.

In particular I'm confused where $n$ comes from and why it's finite, when the cosets may be infinite, and how to construct the set of finite sums from which the infimum is taken. For example, if I start with the simplest possible case by assuming $n=2$ I'm struggling to define $p_2$. It looks like maybe it's saying take the shortest sequence of stepping stones from one coset to the other. The way the indexing of $p$ and $q$ coincides is confusing me. I think an example will help me best. So how do I calculate this pseudometric for my example below?

FWIW I'm aware that under the absolute value metric $d$ the cosets of this quotient contain numbers $x,y$ arbitrarily close to each other and this may yield $\forall [x]\forall [y]:d'([x],[y])=0$ but I still want to understand the metric and calculate it.

Let $M=\Bbb Z[\frac16]^+$ be the set (and multiplicative monoid) of positive rational numbers whose primes greater than $3$ have non-negative powers.

Let us assume for the sake of understanding the quotient pseudometric that $M,d$ is a metric space where $d(x,y)$ is the absolute value metric $d=\lvert x-y\rvert$.

Let $Q=M/\langle2,3\rangle$, i.e. the quotient monoid obtained using the coset of powers of $2$ and $3$: $\langle2,3\rangle=\{2^m3^n:m,n\in\Bbb Z\}$. The sets in $Q$ are of the form $x\cdot\langle2,3\rangle$ where $x$ is some 5-rough positive integer.

What does the quotient pseudometric $d'=d/{\sim}$ look like on $Q$?

For example, what are

  • $d([1],[5])$
  • $d([1],[13])$
  • $d([13],[85])$ and how are they calculated?
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If your metric is based on absolute value, $d(x,y)=|x-y|$, then the pseudometric you get here is trivial, in other words $d([x],[y])=0$ for all $[x],[y]\in Q$. You don't need to use longer and longer sequences of "stepping stones" $[p_i]$. Let's fix an integer $m>0$. Set $$ p_1=x/6^m\in[x]\quad\text{and}\quad q_1=y/6^m\in[y]. $$ Then $d(p_1,q_1)=|x-y|/6^m$. Because we can choose $m$ as large as we wish, this already implies that the infimum is zero.


You need those sequences of stepping stones for things like the following. Let $X=[0,1]$ with the usual metric. Let $\sim$ be the equivalence relation that, for all integers $n>1$, collapses all the intervals of the form $[1/n,1/(n-1))$ to a single point. What's the (pseudo)distance $d([0],[1])$?

  • We have $1/2\sim 99/100$. Consequently, using the sequences $p_1=0$, $p_2=99/100$, $q_1=1/2$, $q_2=1$ to shorten one of the steps: $$d([0],[1])\le d(0,1/2)+d(99/100,1)\le 1/2+1/100.$$ This is less than the original distance $d(0,1)=1$ because we can equate $1/2$ with $99/100$ and instead of $1\to1/2\to0$ we can reroute $1\to99/100, 1/2\to0$.
  • But we also have $1/3\sim49/100$, so with $p_1=0$, $p_2=49/100$, $p_3=99/100$, $q_1=1/3$, $q_2=1/2$, $q_3=1$ we get $$d([0],[1])\le d(0,1/3)+d(49/100,1/2)+d(99/100,1)=1/3+1/100+1/100.$$

By using longer and longer sequences of steps and moving the representatives even closer to the end points of the equivalence classes (i.e. distances less than $1/100$ that I used above), it is easy to see that the infimum over all sequences of stepping stones becomes $0$. In other words, we get $d([0],[1])=0$.