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How do I show a function on 2-adic units is continuous (using the 2-adic metric)?

I'd be happy to learn the general rule or definition. But in particular I need to show that $f(x)=\dfrac{3x+1}{2^{v_2(3x+1)}}$ is continuous at all odd numbers.

Since the function is isometric in $\lvert\cdot\rvert_2$, i.e. since $\lvert f(x)\rvert_2=\lvert x\rvert_2$ every orbit of the function is duplicated, multiplied by any power of $2$. This can therefore be formulated in various ways; as a function though the odd numbers, through the dyadic rationals, but the following seems good to work with. If we let $n$ be a positive integer and $k$ be the power of $2$ we can define:

$f(2^k(2n+1))=(3(2n+1)+1)\cdot2^k\cdot\lvert3(2n+1)+1\rvert_2$

So $f(2^k(2n+1)=(3n+2)\times2^k\times\lvert3n+2\rvert_2$

Here is what I have so far:

I think to prove continuity in the 2-adic metric I need $\lvert x_n-x\rvert_2=0\implies\lvert f(x_n)-f(x)\rvert_2=0$

I think $k$ and $n$ are independent so I think I can examine them independently. Taking $k$ to infinity brings both $x$ and $f(x)$ to zero so that seems to satisfies the continuity requirement.

Moving on to $n$; this seems to be an exercise in proving convergence within odd integers which are in a sense a subset of the 2-adic units. I know all Cauchy sequences in these converge to 2-adic units but not much more than that.

However I do have a little insight into this particular function. For example if we examine the inputs $x$ which map to any given output of $f(x)$, it can fairly easily be shown that these $x$'s take the form of a set $\left\{4^mp+\dfrac{4^m-1}{3}:m,n\in\mathbb{N}\right\}$ so, at least for any given output $f(x)$ the inputs always converge to $x=\frac{-1}{3}$ as the intervals between them become large powers of $2$.

In fact the orbit of the function $a(2^kx)=2^k(4x+1)$ on variation of $x$ and holding $k$ fixed is in a sense orthogonal to $f(x)$; which is a restatement of the above except not in closed form.

Robert Frost
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  • Do you understand how $\mathbb{R}$ is $\mathbb{Q}$ plus the limits of every Cauchy sequences for $|.|$ ? Then $\mathbb{Z}2$ is $\mathbb{Z}$ plus the limits of every Cauchy sequences for $|.|_2$. A function $g : (\mathbb{Z}_2,|.|_2) \to (X,|.|_X)$ is continuous iff $\lim{n \to \infty} |a_n - a| = 0 \implies \lim_{n \to \infty} |g(a_n) - g(a)| = 0$. Thus your function $f : (\mathbb{Q},|.|_2) \to (\mathbb{Q},|.|_2)$ is continuous iff for every Cauchy sequence $a_n \in (\mathbb{Q},|.|_2)$ then $f(a_n)$ is Cauchy $\in (\mathbb{Q},|.|_2)$. – reuns Dec 19 '17 at 20:26
  • @reuns thanks for the hint. Yes I understand completion/Cauchy. Apologies for being daft; is this a) something I should be able to answer from here given your hint, b) too hard, c) obviously false from what you write, or d) provable only if I prove the Collatz Conjecture first!? – Robert Frost Dec 19 '17 at 20:51
  • If you can define $|.|_2$ and prove the addition and multiplication are continuous $(\mathbb{Z},|.|_2)\times (\mathbb{Z},|.|_2) \to (\mathbb{Z},|.|_2) $, then you shouldn't have any problem to answer. Of course replacing $2^{-v_2(x)}$ by $|x|_2$ may help. Do you think $x \mapsto |x|_2$ is continuous $(\mathbb{Z},|.|_2) \to \mathbb{R},|.|)$ ? Is it continuous $(\mathbb{Z},|.|_2) \to (\mathbb{Q},|.|_2)$ ? Maybe more important : do you understand why if $f$ is continuous on $(\mathbb{Z},|.|_2) $ then it is also naturally defined and continuous on $(\mathbb{Z}_2,|.|_2) $ ? – reuns Dec 19 '17 at 21:05
  • @reuns I'm clueless really whether $x \mapsto |x|_2$ is continuous but my GUESS is that for x to converge in Z the differences are ascending powers of 2 so the 2-adic metric function in R is converging to zero, so that IS continuous. But $(\mathbb{Z},|.|_2) \to (\mathbb{Q},|.|_2)$ is NOT continuous since ascending powers of two differences will translate into decreasing powers of $2$ in $\lvert x\rvert_2$ which is diverging. Is that correct? – Robert Frost Dec 19 '17 at 21:11
  • Re your last question, I think continuous $f$ is naturally defined and continuous on $(\mathbb{Z}_2,|.|_2)$ because $Z_2$ is the field defined by sequences which are Cauchy in $\lvert\cdot\rvert_2$ which means $Z_2$ Cauchy sequences can't transcend $Z_2$. But this is the language of a non-mathematician with no access to teachers. – Robert Frost Dec 19 '17 at 21:15
  • @reuns p.s. I believe this function is NOT continuous in many regions of $\mathbb{Z}_2$ and $\mathbb{Q}_2$ but it should be continuous in the locations I specify; namely numbers of the form $(2\mathbb{N}-1)\times{\ldots,\frac{1}{2},1,2,4,8,\ldots}$ so I THINK perhaps I need to use the convergence rule you gave above but only in sequences comprising of these numbers? This would prove the weakened Collatz conjecture (no nontrivial loops). – Robert Frost Dec 19 '17 at 21:22
  • I think you should work on the different definitions of $\mathbb{Z}_2$ : as limits of Cauchy sequences, as $2$-adic series, as sequences $c = (c_1,c_2,\ldots), c_k \in \mathbb{Z}/2^k \mathbb{Z},c_m \equiv c_k \bmod 2^k$ for $m \ge k$, trying to make concrete what means $\mathbb{Z}_2$ and $|.|_2$. – reuns Dec 19 '17 at 21:30
  • @reuns ok but i can't leave yet another conversation with you with no feedback on whether what I wrote was correct. Just to say "yes you're right, $x \mapsto |x|_2$ IS continuous $(\mathbb{Z},|.|_2) \to \mathbb{R},|.|)$" and "yes you're right it's NOT continuous $(\mathbb{Z},|.|_2) \to (\mathbb{Q},|.|_2)$", or the converse, that I'm wrong, would be helpful but I've spent nearly an hour's work to answer these questions you ask and then to repeatedly get no feedback is pretty devastating to be honest. I know you're trying to help. – Robert Frost Dec 19 '17 at 21:36
  • Your last comment is not wrong, but what is your argument for that ? That's where defining precisely an arbitrary element $a \in \mathbb{Z}2$ and a Cauchy sequence $a_n \in \mathbb{Z}$ converging to it, would help. For example $a = \sum{i=0}^\infty b_i 2^i, b_i \in { 0,1}, a_n = \sum_{i=0}^n b_i 2^i$, $|a-a_n|2 = |\sum{i=n+1}^\infty b_i 2^i|_2 \le |2^{n+1}|_2 = 2^{-n-1}$. – reuns Dec 19 '17 at 22:07
  • @reuns ah ok thanks. I can construct sequences of odd integers $x_n$ that converge and $f(x_n)$ as i define it above seems to converge but I can't construct an arbitrary, general statement. – Robert Frost Dec 19 '17 at 22:11
  • What makes you think odd integers help ? – reuns Dec 19 '17 at 22:14
  • @reuns because I understand the only continuous function which cycles in the p adics is the identity function. So this being continuous in the odd integers is compatible with the Collatz conjecture being true, as the only known cycle in the odd integers is the identity function $1\to1$. But the Collatz conjecture has cycles of order greater than one in some rational numbers so this cannot be continuous in the 2 adics in the vicinity of those rationals. – Robert Frost Dec 20 '17 at 00:00
  • @reuns I believe this is closely related to Sharkovski's theorem and that a) some isomorphism exists between some subset of the 2 adics units and a segment of the real line such that this, or some extension of it is a continuous function and therefore the only cycle is the identity map $1\to1$ follows from the fact that any nontrivial cycle would contradict Sharkovski's theorem and b) the function can itself be shown to be continuous in the odd integers, providing a 2nd proof of the weakened Collatz conjecture. – Robert Frost Dec 20 '17 at 00:08
  • Your last comments don't make any sense. The Collatz function $f(2n+1)= 3(2n+1)+1, f(2n) = n$ is continuous $\mathbb{Z},|.|_2 \to \mathbb{Z},|.|_2$ and it contains arbitrary long cycles. It extends naturally to $f : \mathbb{Z}_2 \to \mathbb{Z}_2, f(n) = 3n+1$ if $|n|_2 = 1$, $f(n) = n/2$ otherwise. Why don't you read the books on Collatz conjecture ? – reuns Dec 20 '17 at 03:36
  • @reuns. This prospective proof is only valid when you use the Collatz map $2\mathbb{N}-1\to2\mathbb{N}-1$ as I give above. In the version you describe I doubt continuity at the even numbers equivalent to 1 modulo 3 due to the switching there from $3x+1$ to $n/2$. – Robert Frost Dec 20 '17 at 05:28
  • @reuns p.s. my extension to the 2-adics is different to yours. I am well aware of the extension you give. I think mine treats proper dyadic fractions differently to that, and in fact all numbers with 2 adics valuations greater than 1. – Robert Frost Dec 20 '17 at 05:35
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    ??? Nonsense again. The Collatz function is continuous for $|.|_2$ and has a canonical continuous extension to $\mathbb{Z}_2$. But its condensed version $\tilde{f}(2n+1)= 3(2n+1)+1, \tilde{f}(2^k (2n+1)) = 2n+1$ is not continuous. – reuns Dec 20 '17 at 05:47
  • @reuns ah then perhaps I have misconstrued cyclic for periodic. Here: https://sbseminar.wordpress.com/2009/02/18/there-is-no-p-adic-2-pi-i/ it says there are no nontrivial periodic functions in Qp so is this wrong or have I mistranslated the meaning of periodic? Perhaps what I SHOULD be doing is to restate $f^n(x)$ as a function of $n$ and show that THAT is continuous? – Robert Frost Dec 20 '17 at 09:46
  • @reuns Your "canonical" extension doesn't extend to rational numbers with even denominators while mine does. It's all there in the question. Both functions $g(x)$ and $h(x)$ are quite explicit. I think "nonsense" is a word best used with care. – Robert Frost Dec 21 '17 at 19:55
  • @reuns I may be mistaken but I don't see how what you write as $\tilde{f}(2n+1)= 3(2n+1)+1, \tilde{f}(2^k (2n+1)) = 2n+1$ makes the extension I give in the question not continuous. My function in your language is $\tilde{f}(2^k (2n+1)) = 2^{k-p}(3(2n+1)+1)$ where $p$ can be thought of as a "height" in base 2 and is given by $p=v_2(3(2n+1)+1)$....I'm not totally sure you've grasped the fact that it effectively, goes directly from one odd number to the next, skipping the evens. The divisions by $2$ are translated into the value of $p$ – Robert Frost Dec 27 '17 at 20:23
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    $\tilde{f}(2^k (2n+1)) = 2n+1$ is not continuous because $\lim_{k \to \infty} 2^k (2k+1) = 0$ and $\lim_{k \to \infty} \tilde{f}(2^k (2k+1))$ diverges. The non-condensed Collatz function can be extended to a continuous function on $\mathbb{Z}_2$ and $\mathbb{Q}_2$. – reuns Dec 27 '17 at 23:09
  • @reuns thanks. This is really helpful. I will investigate the effects of having $p$ in my version of $\tilde{f}(\cdot)$ – Robert Frost Dec 28 '17 at 03:21
  • @reuns in the answer below I think I've correctly applied what you did in your above comment to my function to show that this fails to prove it not continuous. But I think this is perhaps not sufficient to show it continuous is it? – Robert Frost Dec 29 '17 at 19:50
  • @reuns I've rewritten considerably clarified, if you still have patience ^^. – Robert Frost Dec 30 '17 at 20:32

1 Answers1

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As for the map

$$f: x \mapsto \frac{3x+1}{2^{v_2(3x+1)}} = (3x+1)\cdot |3x+1|_2,$$

it is the composition of $x\mapsto 3x+1$ and $y\mapsto y |y|_2$, so we want to enquire where these are continuous, the only interesting part being actually the absolute value map $| \cdot|_2$ itself. Viewed as map $(\Bbb{Q}_2, |\cdot|_2)\rightarrow (\Bbb{Q}_2, |\cdot|_2)$, the absolute value is not continuous at $0$ (because $|2^n|_2 =2^{-n}$ does not converge $2$-adically for $n\to \infty$), but outside of $0$, it is actually locally constant and hence continuous. So the composite map $g: (\Bbb{Q}_2, |\cdot|_2)\rightarrow (\Bbb{Q}_2, |\cdot|_2)$ is continuous everywhere except at $x=-\frac{1}{3}$. (Note, however, that this point $-1/3$ w.r.t. the $2$-adic metric does lie in every neighbourhood of $\mathbb{N}$, even in every neighbourhood of the odd natural numbers, as mentioned here recently.)

With a similar argument, the function $\tilde f$ in your answer -- which, if I understand it correctly, is nothing else than $x\mapsto |x|_2^{-1}\cdot f(x\cdot |x|_2)$ -- is continuous as function $(\Bbb{Q}_2, |\cdot|_2)\rightarrow (\Bbb{Q}_2, |\cdot|_2)$, except at the points $-\frac{2^k}{3}, k \in \mathbb{Z}$.

As for the function $g$, which I would rewrite as $x\mapsto (3x+|x|_2^{-1})\cdot |3x+|x|_2^{-1}|_2$, it looks as if it is continuous except at $0$ and all $-\frac{2^k}{3}, k \in \mathbb{Z}$.

Torsten Schoeneberg
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  • Thanks, this is fabulous, I am much clearer now. I edited and simplified the question and deleted my attempted answer so I may adjust those amends in the morning to ensure your answer makes sense. The first thing I am unsure of here, is the fact that to treat the function as a composition of two functions you had to use a function which leaves the odd numbers, and whether this may affect the validity of the composition as a means of deciding the function is continuous in the odd numbers. Is there any doubt over this? – Robert Frost Dec 31 '17 at 02:01
  • P.s. you may be able to show much faster than me that this implies the orbit of $\tilde{f}$ can not be periodic in the integers other than period $1$, which I hope now follows fairly quickly. – Robert Frost Dec 31 '17 at 02:40
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    I took your function(s) as defined on all of $\mathbb{Q}_2$. The continuity results stay the same when you restrict it to any subspace. By the way, one has $|f(x)|_2 = 1$ for all $x \in \mathbb{Q}_2$ except $x=-1/3$; in particular, $f$ is only isometric if you restrict it to a subset consisting of elements of absolute value $1$ (like the odd integers). -- I do not know what it means for a function to have "a periodic orbit", and for my guesses of what that might mean, I do not see how it would follow from those continuity results. – Torsten Schoeneberg Dec 31 '17 at 03:35
  • Ok I will think more but I think this could be an important result. Orbit is just repeated composition. Period $n$ just means the orbit gets back where it started in $n$ compositions so $\exists x|\tilde{f}^n(x)=x$. My thinking is... the orbit of the Collatz function is closed to the odd integers on composition. I have an argument there are no continuous periodic functions in the 2-adic space not of period 1. It may translate to the odd numbers proving no nontrivial loops in the Collatz function. – Robert Frost Dec 31 '17 at 07:52
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    Then that argument you have is wrong. Replace $3x+1$ by $5x+1$ in the formula; this also gives a function that is continuous $2$-adically on the odd integers, but $f(13) = 33, f^2(13) = 83, f^3(13)=13$. (Off-topic advice: It is you who should immediately think of such easy counterexamples, not me. You should develop the habit of checking "wait, if I think this proves Collatz, why does it not work in the variations where there are counterexamples?" This would save you and others frustration and embarrassment.) – Torsten Schoeneberg Dec 31 '17 at 19:45
  • There are other counterexamples too, e.g. many rational numbers. More work is required. The function needs converting into an orbit $\tilde{f}^n(x)$ then that converted to a function on $n$ rather than $x$. I now realise it is the continuity of that as a function on $n$ which needs proving continuous. I'm putting together a nice power series representation which has hope to do that. Furthermore, on another tack, by Sharkovski's theorem if there is an isomorphism from a segment of $\mathbb{R}$ to the 2 adics, continuity as shown above proves the theorem, provided the odd integers... – Robert Frost Dec 31 '17 at 19:59
  • ...can be isolated from the rational and negative numbers which DO cycle. Only the positive integers and selected rationals do not cycle so their image would need to be mapped into a segment of R by the morphism and there would have to be no cyclic points in that segment. However I'm unsure which of Q2 would be included to give us an uncountable set. Regardless I have learnt several useful things about continuity. – Robert Frost Dec 31 '17 at 20:04
  • ... The need to build an uncountable set in Q2 which can be mapped to a segment in R is what motivated my previous question to extend the 2-adic valuation to non-integer values. Then the isometric property of $\tilde{f}$ could be extended using this valuation to include uncountable many isomorphic sequences. – Robert Frost Dec 31 '17 at 20:09
  • ...however it now occurs to me that may not be necessary as there can be uncountably many p-adics with a certain valuation. – Robert Frost Jan 01 '18 at 02:48