What is the quotient metric on this set of equivalent pairs drawn from $\Bbb Q$? Let $X$ be the set of dyadic rationals in the interval $\left[\frac12,1\right)$ which are coprime with $3$, and let the surjection $f:X\to Y$ where $Y\subsetneq \Bbb Q$ be given by
$$f(x)=\begin{cases}\frac{4x}3 &\text{if}& x<\frac34\\
\frac{2x}3& \text{if}& x>\frac34\end{cases} $$
Let $Z=X\cup Y$ inherit the standard (absolute value) topology from $\Bbb R$.
Now define the equivalence relation $\sim$ on $Z$ such that

*

*$\forall x: x\sim f(x)$ and

*$\forall x: f(x)\sim x$ and

*$\forall x: x\sim x$, and

*if $a\sim b$ and $b\sim c$ then $a\sim c$.

Then the set of equivalence classes $P=Z/{\sim}$ comprises elements which are pairs $(x\sim y)$ such that $x\in X$ and $y\in Y$, and forms exact covers of both $X$ and of $Y$.
Question

$P$ has the quotient (pseudo)metric derived from $\Bbb R/{\sim}$ on $x\sim y$.  Is that pseudometric a metric, and what is it equal to?

Attempt
I've read about a somewhat complicated $\sup...$ definition over sequences of class elements as the standard method of defining a quotient metric.  My first instinct is that that's overkill here because every equivalence class here is just a pair.
I thought $d(p_1,p_2)=\min \{d(a,b):a\in p_1, b\in p_2\}$ was the obvious pick - satisfying the metric axioms.  But I was unable to verify the triangle inequality.  Is that right?  Is $d:P\times P\to\Bbb R$ the quotient metric in this case.
Footnote
But also, this quotient space is a setting in which the level sets of the Collatz graph converge to their image.  So I wouldn't be surprised if the more complicated chain definition of the quotient $\begin{equation*} d'([x],[y]) = \inf\{ d(p_1,q_1) + \cdots+ d(p_n,q_n):p_1 = x,q_i \sim p_{i+1},q_n=y\} \end{equation*}$ were related to Collatz sequences.
 A: Not an ideal answer because I don't know what the quotient (pseudo)metric (which I find well-explained in Why are quotient metric spaces defined this way? and Why are quotient metric spaces defined this way?) on $Z/{\sim}$ actually is; however, I can show that your "attempt" is not it, as it is not a pseudometric at all.
Namely, let $x_1 := \frac{17}{32}, x_2 :=\frac{19}{32}$, accordingly $y_1=\frac{17}{24}, y_2 = \frac{19}{24}$. Your attempt would just give distance $d_{attempt}([x_1], [x_2]) =1/16 =0.0625$.
However, now look at $x_3:=\frac{811}{1024}$ with $y_3=\frac{811}{1536}$. The joke is that in the standard metric, $d(x_1,y_3) \approx 0.0033$ and $d(x_3,y_2) \approx 0.0003$ meaning that $d_{attempt}([x_1], [x_3]) + d_{attempt}([x_3], [x_2]) \approx 0.0036$ which is much smaller than $0.0625$, i.e. the triangle inequality is violated. Indeed, with the standard definition of the quotient pseudometric (see links above), this already shows that the quotient pseudodistance between $[x_1]$ and $[x_2]$ is smaller than $0.0036$.
Similar tricks (the idea of the above is that "up to $\varepsilon$, we can identify $x_2 \sim \frac23 \cdot\frac43\cdot x_2 = \frac{8}{9}x_2$") work for other numbers, and my suspicion would be that one can ultimately use numbers of the form $\frac{2^n}{3^m}$ arbitrarily close to $1$ to show that the quotient pseudometric is actually identically $0$; but I admit I have not really thought through that rigorously. The above example might be a starting point for the interested.
