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Let $M=\Bbb Z[\frac16]^+$ be the multiplicative monoid of positive dyadic and ternary rationals.

Let $Q=\Bbb Z[\frac16]^+/\langle2,3\rangle$ be the quotient that sets $x\sim y\iff\exists p,q\in\Bbb Z:2^p3^qx=y$ and having cosets $[x]\cdot\langle2,3\rangle$ where $[x]\in S$ the 5-rough integers.

Then let $[\space]:M\to S$ send any integer to its unique 5-rough representative.

Let $\phi^n(x)=x+(1-2^{-6n})\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$.

Since $2^p3^q\phi^1(x)=\phi^1(2^p3^qx)$, it follows that $\phi$ acts faithfully on the quotient $Q$.

Since $\phi^{a+b}(x)=\phi^{a}(\phi^{b}(x))$ it follows that $n$ has a monoidal action over $Q$.

Since every sequence $\phi^n:n\in\Bbb N$ is well-founded, $n$ does not have a group action over either $M$ or $Q$. Therefore it follows from this that $\phi$ cannot be a homeomorphism, even before I define any topological space.

Since $[\lim_{n\to\infty}\phi^n(x)]=c(x)$ where $c(x)$ is the first 5-rough successor of any 5-rough integer $x$ in the Collatz graph, it follows that $[\lim_{n\to\omega^2}\phi^n(x)]=1$ is equivalent to saying all 5-rough positive integers converge on iteration of the Collatz function.

Since $2^m3^n[x]:n>0$ are the leaves of the Collatz graph it follows that the 5-rough positive integers are a sufficient set so this is equivalent to the Collatz conjecture.

Since $\lim_{n\to\omega}\phi^n$ surjects over both $M$ and $Q$, it follows that $\lim_{n\to\omega}\phi^n$ is an epimorphism.

The kernel of $\phi^\infty$ is $\{\phi^n\{1,5\}:n\in\Bbb N_{\geq0}\}$

Now let $g:Q\to Q$ and $g(x)=[\lim_{n\to\omega}\phi^n([x])]\langle2,3\rangle$

Note that the square brackets pick out the 5-rough representative but they are really quite pointless as the function acts faithfully on the quotient irrespective of the representative chosen, so one could equivalently just write $g(x)=\lim_{n\to\omega}\phi^n(x)$.

What is the finest topology on $Q$ such that $g$ is continuous?

My proposed answer is to let the infinite sequences $\phi^n(x)$ be the open sets

Let $\tau=\{\{\phi^n(x):n\in\Bbb N\}\cdot\langle2,3\rangle:x\in Q\}$ be the open sets of $Q$.

Then $Q,\tau$ is the finest such topology, but I am very inexperienced in such matters so please bear with me... This question may just be about clarifying misconceptions on my part.

Does the nature of the kernel, the function $\phi$ and the fact that $\langle2,3\rangle$ is easily proven to be the sole fixed point of $Q$, preclude the existence of cyclic sets?

An interesting point to note is that there is a function $h$ from $Q$ into a subset of $\omega^\omega$ such that $h\lim_{n\to\omega}\phi^n(h^{-1}x)$ acts transitively if and only if the Collatz conjecture is true.

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    $\begingroup$ I'm not sure that's the question you want to ask. The finest topology that makes $g$ continuous is of course the discrete topology, since it makes every function continuous, and it is simply the finest topology. $\endgroup$ – Captain Lama Nov 24 '19 at 0:27
  • $\begingroup$ @CaptainLama ok thanks. $\endgroup$ – samerivertwice Nov 24 '19 at 3:22
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If $S$ is any set and $f: S \rightarrow S$ is any function, the finest topology which makes $f$ continuous happens to be the finest topology that $S$ can be endowed with, i.e. the discrete topology.

This is a special case of the fact that if $f:X \rightarrow Y$ and $Y$ is any topological space, then the finest topology on $X$ which makes $f$ continuous is the discrete topology.

In even greater generality and a little vaguely, if we are free to choose topologies independently on both $X$ and $Y$ for a map $f:X \rightarrow Y$, the map $f$ has a higher chance to be continuous the coarser the topology on $Y$, and the finer the topology on $X$ is.

All this is made more precise, via the notions of "inital" and "final" topologies, in any introductory course on topology.

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  • $\begingroup$ I do not know. - $\endgroup$ – Torsten Schoeneberg Dec 3 '19 at 19:32
  • $\begingroup$ No, that would change nothing about my answer. $\endgroup$ – Torsten Schoeneberg Dec 3 '19 at 22:57
  • $\begingroup$ I've unaccepted this answer for the time being because I think it would be significantly clearer if it emphasised that if $f:X\to Y$ then this answer is dependent upon the predicate $X=Y$. $\endgroup$ – samerivertwice Jan 2 at 21:23
  • $\begingroup$ There you go. Of course nothing in your original question asked about that; you had the same set as domain and codomain and asked for one topology on that set, so I and maybe five other people found my original answer quite elegant, being as it was obviously true without even looking at whatever $f$ and whatever half-sensible definition of its domain-codomain $G$ you had come up with for the time being. $\endgroup$ – Torsten Schoeneberg Jan 2 at 21:44
  • $\begingroup$ Thanks. I'm not judging you or the answer, just following site rules on voting. Hopefully this will now be clearer in the unlikely event anyone comes along here in the future with a similar query. I initially accepted your answer without being any the wiser, assuming it was just I being dumb. The significance of $S\to S$ only became apparent to me upon reading Paul Frost's answer to my subsequent question. $\endgroup$ – samerivertwice Jan 2 at 22:18

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