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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.

1 Answers1

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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.

  • Thank-you very much. I guess a good next step is to take the two Collatz sequences from any two given numbers to $\frac12,\frac23$ and try to prove the sum of those chains is (or is not) the quotient metric. What we have on our side is that we know $\frac12,\frac23$ is the only fixed point of the function. The other "quick" attempts at metrics I guess are $\sup$ and $\inf$ but treating the pairs as elements of $c$ – it's a hire car baby Sep 05 '20 at 09:26
  • Re your last point; that this may give the trivial metric $d'=0$: Clearly we need infinitely long chains to get to $d'=0$. But is there some way to get to a topology with a finite distance? e.g. every point is within distance $d_n$ of $\frac12,\frac23$ within $n$ steps? In the Collatz graph, every chain of $x\mapsto x+\frac132^{\nu_2(x)}$ gets to $0$ (2-adically) in infinitely many steps. But for all positive integers it always gets to $\langle2,3\rangle$ in finite steps. So this is the thing I want to capture. The quotient space in this question is designed to do that (but probably misses). – it's a hire car baby Sep 07 '20 at 09:18
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    In general one does not need arbitrarily long chains for the quotient pseudometric to be $0$ (example: Look at $\mathbb R/ \mathbb Q$, where already the infimum of added distances over chains of length 3 is $0$). If that is the case in this example I do not know, meaning I do not see your "Clearly, ...". – Torsten Schoeneberg Sep 07 '20 at 22:21
  • ok, thanks for thinking about it. I'll think about that $\Bbb R/\Bbb Q$ example as I can't see it... we may mean different things by "chains". But also i guess those classes are infinitely large. The classes in this question are all just pairs. – it's a hire car baby Sep 08 '20 at 00:12
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    By "chain" I meant the sequences $p_1, ..., p_n$ resp. $q_1, ... q_n$ in the linked definitions of quotient pseudometrics, i.e. by "length" thereof, I meant $n$. If you meant something else, obviously my comment is void. – Torsten Schoeneberg Sep 08 '20 at 00:32
  • I got it after I had a think about what $\Bbb R/\Bbb Q$ looked like. The case in this question is somewhat different with every class being only a pair and constructed to only step one change of power of $3$ per step, then I think a chain at least long as $\lvert\nu_3(a-b)\rvert$ is needed here to span $a,b$. Or something like that. – it's a hire car baby Sep 08 '20 at 00:49
  • Is it obvious whether this quotient space is connected? – it's a hire car baby Sep 10 '20 at 11:39
  • In case you're interested I received an answer to this on MO. It turns out there's an elegant way of showing the orbits of $f$ are topologically conjugate to an irrational map on the unit circle. That ensures they're dense on the unit circle, so distances are zero. https://mathoverflow.net/a/380398/91341 – it's a hire car baby Jan 11 '21 at 22:08
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    Good for you. I see the basic idea there follows along what I had suspected here, but the trick of translating to translation by $\mathrm{log}_2(3)$ via exponents was invisible to me and is very neat. -- It would have been nice of you, though, to notify both there and here that this was a crosspost between MSE and MO. – Torsten Schoeneberg Jan 14 '21 at 07:57
  • I didn't think it a crosspost at the time I posted it because I thought the restriction in the MO question to equivalence classes of cardinality two made it a fundamentally different question. It was only the answer to that one which revealed to me that that proof for equivalence classes of cardinality two generalises this question too. – it's a hire car baby Jan 14 '21 at 15:39