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.

  • 3
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.