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1:36 AM
 
 
12 hours later…
1:43 PM
@samerivertwice There are several comments in reply to your question which likely explain the downvotes:
Your resoluteness is unfounded. I have pointed out specific issues and asked specific questions which you have not addressed. Your "best candidate" as written ($x,y$ being cosets) makes no sense for the obvious reason that none of the operations in your formula is defined on those cosets. If one tries to save it by choosing representatives, it does not even satisfy $d(x,x)=0$, nor symmetry, and I have no motivation to check the triangle inequality. You might say you mean something else, but the issue is that no-one, including you, can clearly state what you might mean. — Torsten Schoeneberg 5 hours ago
In your head you are most likely switching between cosets and representatives in the middle of calculations. Ca. a year ago I put considerable effort into an essay-like answer to one of your questions, explaining how wrong that is; that question and answer unfortunately have been deleted. Back then you thanked me and promised to think about it all deeply; now you are making the same basic mistakes again, showing you have no understanding of almost any mathematical term you use so lavishly. I am not at all able to illuminate you. Maybe ask one of the people who voted to reopen this question. — Torsten Schoeneberg 5 hours ago
 
2:38 PM
@XanderHenderson Hi Xander, thanks for the ping. In my view, the most likely scenario is that Torsten has either a) failed to satisfy himself that $f^n(x)$ commutes with $2^p3^q$, b) failed to satisfy himself that it surjects, or c) failed to appreciate that commutativity makes it well-defined on $\Bbb N/\langle2,3\rangle$. It'd be a shame if multiple downvotes followed from his comments because his claim that my own attempted metric is flawed, is after all the primary premise of the question.
 
 
4 hours later…
RRL
6:16 PM
Delete: DA, DB, DC, DD, DE, DF, DG, DH
 
6:54 PM
@samerivertwice Torsten Shoeneberg has demonstrated that the thing you claim is a metric fails to be a metric in basically every way imaginable. You claim you know this. Fine. Then why are you calling it a metric?
He has also pointed out that you are performing operations on representatives, and are claiming that these are operations on equivalence classes. Do you know that these operations are well-defined on equivalence classes?
Finally, you ask answers to define, for you, a metric which has the property that $f^n$ converges.
How do you know that such a metric exists? It is certainly possible that you may define a topology such that these sequences converge, but it is not immediately obvious that (1) any such topology would be unique, nor (2) that any such topology would necessarily be metrizable.
In short, there are a number of issues with your question, which more-or-less reduce to the statement "Your question is unclear."
As such, I am sympathetic to the downvoters.
I might further argue that your question is lacking context: what do you mean by a "5-rough representative of $x$"? in what sense do you mean that $f^n : n < \omega$ is "well-founded"? and so on.
Presumably, these are ideas understood by your intended audience, but it can't hurt to be more explicit.
 
7:36 PM
@XanderHenderson thanks for the note. You ask why I include $d$ when I know it's not a metric. This is to show my own (failed) attempt as per the site policy you and others so strictly police.
@XanderHenderson the function $f$ commutes under multiplication with powers of $2$ and $3$ in the sense that $2^p3^qf(x)=f(2^p3^q(x)$, and since the quotient is defined multiplicatively the same function is equally perfectly defined on the master monoid/group, as it is on the equivalence classes. So Torsten's claim the function operates on representatives is false. Maybe this is what I should clarify.
So he is not so much pointing it out, rather misunderstanding what is going on. I will review and see if it is my wording that has misled him.
 
8:08 PM
@samerivertwice Then the fundamental issue is that your question is unclear.
 
@XanderHenderson well-founded simply means there are finitely many predecessors, in this case I'm referring to the action of $n$. I will clarify.
 

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