# What is the finest topology that makes $\lim_{n\to\infty} x+(1-2^{-6n})\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$ continuous?

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.

• 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. – Captain Lama Nov 24 '19 at 0:27
• @CaptainLama ok thanks. – 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.

• I do not know. - – Torsten Schoeneberg Dec 3 '19 at 19:32
• No, that would change nothing about my answer. – Torsten Schoeneberg Dec 3 '19 at 22:57
• 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 $X=Y$. – samerivertwice Jan 2 at 21:23
• 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. – Torsten Schoeneberg Jan 2 at 21:44
• 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. – samerivertwice Jan 2 at 22:18