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I've been studying the topology of a certain form of the Collatz graph with its topology inherited from $Bbb Q_2$ and I've managed to determine that the kernel of the Collatz function contains as its limit point its supposed destination - the number $1$.

Furthermore with a little work I think it can be shown the entire graph is bounded by $1$.

With still fairly rudimentary knowledge of compact metric spaces, I'm somewhat limited in knowing what, if anything, can be deduced from this - although intuition tells me there is something.

Let $G=Bbb Z[frac16]setminus{{0}}$

Let $N$ be the multiplicative group generated by the elements: $-1,2,3$

Then let $X=G/N$, i.e. the set of multiplicative cosets $X={gN:gin G}$. Now let $X$ inherit its topology from $Bbb Q_2$ and note that every coset can easily be identify by its unique representative $x^times$ satisfying $lvert x^times rvert_2=lvert x^timesrvert_3=1$

The Collatz conjecture states that the following surjection $f:Xto X$ stabilises finitely to $N$ for all inputs:

$f(x)=3x+text{sign}(x)cdot2^{nu_2(x)}cdot3^{nu_3(x)}$

If wanted we can add $f(0)=0$ and retain continuity.

The preimage in $f$ of every singleton in $X$ has cardinality $aleph_0$ and the kernel of $f$ defined as $ker(f)={gin G:f(g)in N}$ is given by:

${1,5,85,341,5451,ldots}cdot N$

in which the indexing elements are composed of the pair $x_1=1,x_2=5$ and the recurrence relation $x_{n+2}=g(x_n)$ where $g(x)=64x+text{sign}(x)cdot21cdot2^{nu_2(x)}cdot3^{nu_3(x)}$ which is a representative to representative function. The limit point of the orbit of $g(x)$ is $-frac13cdot N$ which is of course in $N$, the set which the conjecture requires proof of convergence to.

What significance, if any, can be put upon this fact? Not just the inclusion of $N$, which simply follows from $Nmapsto N$ being the sole fixed point of this formulation, but specifically of it being the sole limit point of the kernel.

Furthermore, it is certain that the preimage in $f$ of every singleton in $X$ also has boundary $N$.

Moreover, some basic testing of certain cases shows that it may be the case that:

$forall ninBbb N:$ the preimage in $f^n$ of any singleton also has $N$ as its limit point.

What might be needed to show that this "lateral" bound to $f$ translates into a "forward" bound under finite iteration?

(As a potentially interesting aside, orthogonal to the Collatz function $f$, the graph on $Bbb Z[frac16]setminuslangle3ranglecdotBbb Z[frac16]$ has a Cantor-set-like topology arbitrarily close to $[1]$). The more precise statement is as follows: Let $x_{m+1}=64x_m+text{sign}(x_m)cdot2^{nu_2(x_m)}cdot21$ define a sequence of integers $S$ such that every $sin S$ has the same immediate successor $[x]$, i.e. satisfying $d_2([x],s)=1forall sin S$. Then $S$ is one of three subsequences of the sequence $S^{frac13}$ given by $x_{m+1}=4x_m+text{sign}(x_m)cdot2^{nu_2(x)}$. But it is always the case that every third subsequence is in $3cdot[x]$ and therefore has a different successor and is not in the same vicinity by $d_2$.

Here's a picture showing the topology of the unique representatives of the cosets (sorry about the neck-bending - I don't know how to make imgur not rotate it onto its side):

This is not an answer, just an attempt to track down where the thinking in the question (during its many reworded versions, which makes it hard to follow) goes wrong.

What we have: $G$ is a monoid, $N$ is a submonoid and even a subgroup. We can form the quotient set $G/N$.

One set of representatives for $G/N$ is given by $lbrace k in Bbb N: 2,3 ; notvert k rbrace$, a set which I will call $Y$.

You have a function $f(x)=3x+text{sign}(x)cdot2^{nu_2(x)}cdot3^{nu_3(x)}$ on $G$.

It is indeed true that $f(yn) = f(y)n$ for all $y in Y, n in N$. This means that $f$ induces a map

$f^* : G/N rightarrow G/N$

It is however noteworthy that:

• $f$ does not respect the monoid structure of $G$, and $f^*$ does not respect the monoid structure on $G/N$. In particular, the preimages $f^{-1}(1)$ resp. $f^{* -1}(N)$ should not be called "kernel" of the respective maps, as this could mislead one into thinking that e.g. $f(g cdot f^{-1}(1) )= f(g)$ which is wrong for most $g$.




• $f$ is not continuous w.r.t. the $2$-adic metric.

Now the next thing I would write is that $f^*$ is also not continuous, but that brings me to one heart of the issue: I do not see any succesful attempt to even define a meaningful topology (or even metric) on $G/N$. This is the deeper issue I want to explicate: Through your edits, you have sometimes called the set of representatives $X$, sometimes you have called $G/N$ $X$. You don't seem to understand the necessity to distinguish between them. Well, they are in bijection, and don't mathematicians identify such things all the time? Yes, because we have learned through practice how far we are allowed to take these identifications and where to stop, in each specific situation.

How far can one take it? Well, there certainly is a set bijection (that's basically what the concept of "representatives" means). But nobody is just interested in a set as set; what is interesting is structures on such a set, like addition, multiplication (algebraic structures), metric, topologies (topological structures), order, etc., and their interrelations / compatibility.

The set $G$ has plenty of such structures and interplay between them. But if one forms a quotient $G/N$, one has to check which of those structures still make sense on the quotient (maybe can even be defined purely by referring to some representatives) and which of them don't.

Well in this case, exactly because $N$ is a (normal) subgroup, the monoid structure transports well to $G/N$. Which just means that one can compute a product in $G/N$, say, $g_1 N cdot g_2N$, by just multiplying their representatives $y_1$ and $y_2$ in $Y$.

So far so good, and this is an example of what I mean by a structure which is compatible with taking the quotient (or sometimes mathematicians say, there is a "natural" or "induced" structure on the quotient which "comes from" a corresponding structure on the big set). Maybe you took that for granted; which one should not. It would not work for just any set $N$. It is worth thinking about this.

So far, no other structure which $G$ exhibits has been shown to induce some like structure on the quotient $G/N$.

One of your recurring fallacies is that you believe you can just use any structure which you have on some set of representatives (viewed as subset of the original $G$) as a structure on the quotient $G/N$. In several of your posts you have a similar error where you want to transport the order structure on some interval to its image in $Bbb R / Bbb Z$ which is nonsense -- or more precisely, and maybe that is why this is so hard to accept: viewed in and of itself, it does make (very limited) sense. Using the bijection, one can "transport" any structure that one has on the representatives to the cosets. The issue is that such a transported structure has no reason to be "natural" as above, to be independent of the representatives you choose, and in any case has no reason to be compatible with any other structure defined in any way ("naturally" or not), even if your original structures were compatible with each other.

Two examples for this: Say, you are looking at the quotient of additive groups $Bbb R / Bbb Z$. Every element of this quotient has a unique representative in $[0,1)$. Of course, on these representatives seen as a subset of $Bbb R$ you have the usual order $le$ which satisfies all the nice properties one demands of a transitive ordering compatible with the additive structure. You can say that $1/4 le 1/2$.

But now the error you make is that you think you can just transport that order on the representatives to the quotient $Bbb R/ Bbb Z$ and have an order on it. But do you? Yes, but no. Yes: In and of itself, you now have ordered the cosets. The coset represented by $1/4$ is $le$ the one represented by $1/2$ etc. But No: Whereas on the original set, that order was (for example) compatible with addition, your new order on the cosets, just defined by those representatives, is not. Here for example, you would want that $1/4 le 1/2 Rightarrow 1/4 + 1/2 le 1/2 + 1/2$, but in the quotient, the latter would read $3/4 le 0$, whereas the transport from the representatives you just made demands $0 le 3/4$. So your "transport of structure" lost everything that was good about the structure beyond its sheer existence, e.g. the property $a le b Rightarrow a+c le b+c$ for all $c$.

Another example: The quotient $Bbb Z / 12 Bbb Z$ has standard representatives $lbrace 0,1, ..., 11rbrace$. But you can not just treat that quotient as if it is those twelve numbers seen as elements of $Bbb Z$. For example: You can not say that an element of the quotient is divisible by 5 if its representative is, i.e. if it is 0, 5, or 10: because in the quotient, e.g. $2 =5cdot 10$ hence is divisible by 5. Whereas the property "divisible by 2" happens to match the one on the representatives (that one does go over "naturally"). Or: You cannot say that in the quotient, $0 le 1 le 2 le ... le 11$, or to be very precise: Of course you can say that and call it an order, but it does not have the properties which would make an order useful (and thus be expected by any mathematician as implied meaning if you speak of orders here), namely, it would, just as in the other example, not be compatible with the additive nor multiplicative structure ($2 le 3$ but $8 cdot 2 ge 8 cdot 3$; $4 le 7$ but $4+6 ge 7 +6$).

Now back to this question.

Here, you think that you can transport topological structure on $Y$ (viewed as subset of $G$ viewed as subset of $Bbb Q$ with the $2$-adic metric) to $G/N$ when it seems useful; but then, once you think you need some other result, you implicitly think of a different topological structure.

E.g. you speak of "closures" of certain sets in $G/N$. The concept of "closure" refers to a topology. But what topology on $G/N$ are you talking about? You seem to think for a long time that you can do

Possibility 1: Use the $2$-adic metric on $Y$ viewed as subset of $Bbb Q$. Transport it to $G/N$ via the set bijection between them. That gives a metric on $G/N$. Consequently points are closed.

That is exactly that naive "transport of structure" which I warned about in the examples above.
Well yes, it would give a metric on $G/N$ (a set in which, by the way, $N =1N$ is a single point). In that metric, the distances of $1N,5N, 85N, 341N$ from the point $1N$ are: $0,1/4, 1/4, 1/4$, etc.; as you see, it stays $1/4$, hence $N$ is not in the closure of that set.

No mathematician would even try this metric because one knows that defining something on cosets which depends on the choice of the representatives is doomed to fail. What mathematicians would try here is rather

Possibility 2: The quotient topology from $G$.

I have thought little about how this topology looks like. However, I have a feeling it would not give what you seem to want either. The first question one would ask is: is $N$ closed in $G$? If it is, then points in $G/N$ are closed.

However, in this case, if $N$ is contained in any closure of preimages of $gN$ for $g notin N$, then $f^*$ is not continuous (because preimages of points must be closed, but cannot contain $N$, since $f^*(N) =N$ .

One more point: you sometimes talk about preimages of sets under the map $f$ again; I admit I have lost motivation to sort this out; let me just say that a preimage of the set $gN subset G$ under $f$ is not the same as a preimage of the singleton $gN in G/N$ under the map $f^*$; of course they are related, but if for example the mutual topologies are not fine-tuned to each other, the closure of $f^{-1}(gN)$ as subset of $G$ might have lost all relation to the closure of $f^{* -1}(gN)$ as subset of $G/N$; you cannot switch between them and use something about one and then sometimes of the other.

And one final, unrelated point: In some edits, you insist that $G$ is a subspace of a compact space, as if that might help with something somehow. (Maybe even a flawed "deus ex machina" argument as outlined here .) Well it almost certainly will not, not only because your maps are not continuous so there's no hope of invoking topological arguments anyway, but also: Look up "compactification" and realise that most spaces are subsets of some compact space.