# How do I show a function on 2-adic units is continuous?

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.

• 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 $\in (\mathbb{Q},|.|_2)$. Commented Dec 19, 2017 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!? Commented Dec 19, 2017 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 $(\mathbb{Z}_2,|.|_2)$ ? Commented Dec 19, 2017 at 21:05
• ??? 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. Commented Dec 20, 2017 at 5:47
• $\tilde{f}(2^k (2n+1)) = 2n+1$ 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 $\mathbb{Q}_2$. Commented Dec 27, 2017 at 23:09

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 $-\frac{2^k}{3}, k \in \mathbb{Z}$.

• 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? Commented Dec 31, 2017 at 2:01
• 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. Commented Dec 31, 2017 at 2:40
• 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. Commented Dec 31, 2017 at 3:35
• 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. Commented Dec 31, 2017 at 7:52
• 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.) Commented Dec 31, 2017 at 19:45