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Create an explicit extension of the 2-adic metric to $\mathbb{R}$ in which at least $(1,\infty)$ is connected, $\lvert x\rvert\lvert y\rvert=\lvert xy\rvert$ and $|x+y|_2\leq\max\{|x|_2,|y|_2\}$.

Here is what I've tried so far:

This question discusses an extension of the 2-adic metric to the real numbers and states that one exists, and is furthermore used in Monsky's theorem - but doesn't define it.

Call this metric $\lvert\cdot\rvert_{\times}$

It seems to me the logical extension of the metric to e.g. $\mathbb{Q}_2[\sqrt{2}]$ is simply to allow $$\lvert\cdot\rvert_{\times}=2^{\tfrac{1}{2}}\text{ or possibly }2^{\pm\tfrac{1}{2}}$$ as well as integer powers of $2$.

I may be mistaken. And I'm unsure how to check whether this extension contradicts the required conditions. Perhaps by continuing in this direction all the real numbers can be connected?

it's a hire car baby
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2 Answers2

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Several things in your post seem to be based on misunderstandings.

1) What you call $2$-adic metric is usually and should be called the $2$-adic absolute value. This value $| \cdot |_2$ is then used to define what should really be called "$2$-adic metric", $d(x,y) = |x-y|_2$.

2) A value that satisfies your properties is nonarchimedean, the metric it defines is an ultrametric: $d(x,y) \le \max \{d(x,z), d(z,y)\}$. Every ultrametric space is totally disconnected, meaning that no subset of it which contains more than one point is connected. This is a nice little exercise in ultrametrics as well as something that is covered in any book about this subject.

This shows that what you are asking for is impossible. I would like to point out some further things that I am not sure you are aware of:

3) The $2$-adic absolute value is, first of all, defined on $\mathbb{Q}$. The completion of $\mathbb{Q}$ w.r.t. it is $\mathbb{Q}_2$. The $2$-adic absolute value naturally extends to $\mathbb{Q}_2$.

4) $\mathbb{Q}_2$ and $\mathbb{R}$ are very different fields. Neither of them is in any way contained in the other. One should think of them as going off in different directions from $\mathbb{Q}$.

5) To qualify the last point, some subsets of $\mathbb{Q}_2$ and $\mathbb{R}$ can be "matched", although already here one should be careful. As long as a number $\alpha$ is algebraic over $\mathbb{Q}$, there is a natural extension of the $2$-adic absolute value both from $\mathbb{Q}$ to $\mathbb{Q}(\alpha)$ and from $\mathbb{Q}_2$ to $\mathbb{Q}_2(\alpha)$ (EDIT: ...and up to $[\Bbb Q(\alpha):\Bbb Q]$ reasonable extensions of $|\cdot|_2$ to $\Bbb Q(\alpha)$, namely, one for each non-archimedean place of $\Bbb Q(\alpha)$ lying above $2$ (= one for each irreducible factor of the $\Bbb Q$-minimal polynomial of $\alpha$ over $\Bbb Q_2$; = one for each distinct prime factor in the prime factorisation of $(2)$ in the ring of integers of $\Bbb Q(\alpha)$. In the example $\alpha=\sqrt{2}$, there is only one such extension of the value, as the prime 2 totally ramifies.

As in your example, of course the only reasonable extension of the absolute value to $\alpha = \sqrt c$ for $c \in \mathbb{Q}_2$ is $|\sqrt c| := \sqrt{|c|}$. Much more generally, an element $\alpha$ whose minimal polynomial over $\mathbb{Q}_2$ has degree $d$ is given the value $|N_{\mathbb{Q}_2(\alpha)|\mathbb{Q}_2}(\alpha)|^{1/d}$. This is treated extensively in any book on $p$-adics, local fields etc.

6) In this way, one could explicitly extend the $2$-adic value from $\mathbb{Q}$ (not $\mathbb{Q}_2$, which is not contained in $\mathbb{R}$!) to the algebraic closure of $\mathbb{Q}$ in $\mathbb{R}$. But to extend it further (as in the theories you quote), one sort of arbitrarily extends it to transcendentals. There is absolutely no meaningful way to assign a $2$-adic value to the real numbers $\pi$ or $e$ or any other transcendental number. To be more clear, the axiom of choice guarantees that we can assign $2$-adic values to all of them, but none of our choices would be better than others. You can make them have whatever value you want, as long as they are compatible (in the sense that if $e^\pi$ would happen to be an algebraical expression in, say, $\pi, e$ and some other transcendental number, then that relation would restrict our choices). So to clarify: There is no explicit construction of such a value -- not because there is no such value, but because there are far too infinitely many, and none of them is more reasonable than all others. (Pick some transcendental number. You can choose it to have whatever absolute value $>0$ you like to begin with, but then when you extend it further, you would first have to make a vast number of compatible choices (to everything that is algebraic over that number), but even after that you still have continuum many transcendentals left. Again you pick one and assign some arbitrary value to it, then go on ...! And you can never explicitly list all these possibilities, actually, you have to make continuum many choices, so that inductive process would never even come close to doing it all.)

Torsten Schoeneberg
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  • Thanks on all points. The "arbitrary" assignment of values to transcendental numbers may be sufficient for my purposes and I kind of see how it's impossible to do this explicitly. You may understand my motivation for such a metric better with reference to this answer: https://mathoverflow.net/questions/286234/ Perhaps the extension you describe in part 5) of your answer can be used to do what that requires. – it's a hire car baby Dec 13 '17 at 15:09
  • I do not fully understand what you want there, but I stress point 2). If you want anything that is more than a point to be connected, $p$-adic values are not the way to go. One of the few things one can say about those arbitrary, non-explicit extensions of the value is that in the metric/topology that they define, the only connected subspaces are empty or singletons. – Torsten Schoeneberg Dec 13 '17 at 18:57
  • I's like to probe at your statement 4 please that neither is in any way contained in the other. My understanding is that a field is an extension of a ring and contains elements and operations. My understanding is that $\mathbb{R}$ does contain the elements of $\mathbb{Q}_2$ and more (eg both sets contain $\pi$), but when we map them in the obvious way there is no field isomorphism relating the operations, so it's not contained as a field. Am I mistaken in this regard or is that a fair statement? Or are you saying there additional elements in $\mathbb{Q}_2$ not in $\mathbb{R}$? – it's a hire car baby Dec 19 '17 at 03:09
  • @RobertFrost: In an answer to another post of yours, I linked to this question where they show that e.g. $\mathbb{Q}_2$ contains a square root of $-7$. $\mathbb{R}$ does not contain such an element, as you certainly know. Conversely, $\mathbb{R}$ contains a square root of $2$, whereas $\mathbb{Q}_2$ does not. Also, in what way do you think $\pi$ is contained in $\mathbb{Q}_2$? Which element of $\mathbb{Q}_2$ would that be, and why? – Torsten Schoeneberg Dec 19 '17 at 03:46
  • Thanks. The significance of $\sqrt{-7}$ hadn't clicked until now. As for $\pi$ I thought somebody told me in another question that it's in $\mathbb{Q}_2$ but I can't show it for myself or remember their precise statement. Perhaps they meant some analogue of $\pi$ or a reference to $t$. Nevertheless, the point was transcendental numbers and I guess $t$ is transcendental and we have $\log(2)$ but I take your point it seems the sets are coincident to some degree but not the relations between their elements. – it's a hire car baby Dec 19 '17 at 12:55
  • Thanks again for this. I just re-read the bit where you covered the valuations and see it covers my new query. Less of it now goes straight over my head than did two years ago. I'll try not to miss my 6 monthly repeat in the summer ;) – it's a hire car baby Dec 10 '19 at 12:59
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Although @TorstenSchoeneberg is right in the 6 points of his answer, I just wanted to point out that it is possible to extend any $p$-adic valuation to $\mathbb{R}$ even to $\mathbb{C}$. The construction is as follows: since there exists a field isomorphism $T:\mathbb{C}\to\mathbb{C}_p$ that extends the identity map in $\mathbb{Q}$, you can define the valuation $|\cdot|_*:\mathbb{C}\to\mathbb{R}$ by $|x|_*=|T(x)|_p$. It is not hard to see that $|x+y|_*\leq\max\{|x|_*,|y|_*\}$ for all $x,y\in\mathbb{C}$. Then, the restriction of $|\cdot|_*$ to $\mathbb{R}$ is a valuation on $\mathbb{R}$ that extends the $p$-adic valuation of $\mathbb{Q}$. Of course, this construction cannot be done explicitly since there is not an explicit formula for $T$. We only know about its existence. For a proof of the fact that $\mathbb{C}$ and $\mathbb{C}_p$ are isomorphic you can see page 83 of the book: Non-Archimedean Functional Analysis - [A.C.M. van Rooij] - 1978.

Chilote
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  • Thanks. This may be exactly what I'm looking for. I was perhaps wrong to use the word 'explicit' when I should have said 'as explicit as possible'. Do you have any opinions on: is an explicit formula for $T$ impossible to realise, or just unknown at this point? And by this metric space $\mathbb{ R}$ would be disconnected? Just to check I understand you both correctly. – it's a hire car baby Dec 18 '17 at 06:46
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    Is there an explicit formula for $T$ impossible to realize? I do not know. If you see the proof that I cited in the answer, all depends on whether we are able to match explicitly two sets with the cardinality of $\mathbb{R}$. So in my opinion, no. But an opinion is not always is a fact. Regarding the disconnectedness of $\mathbb{R}$ under the (ultra)metric $D(x,y)=|x-y|_*$, @TorstenSchoeneberg is right. Any subset of the metric space $(\mathbb{R},D)$ of at least two elements is disconnected because the strong triangle inequality: $D(x,y)\leq\max{D(x,z),D(z,y)}$. – Chilote Dec 18 '17 at 15:43
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    @RobertFrost: How can this be "exactly what you're looking for" if it's an elaboration of what I said -- that assuming the axiom of choice, there are (uncountably many) extensions of the $p$-adic value to $\mathbb{R}$ (and $\mathbb{C}$), but they are non-explicit since essentially arbitrary and thus meaningless on transcendentals? Compare Asaf Karagila's answer here for why such isomorphisms $T$ cannot be made explicit. – Torsten Schoeneberg Dec 19 '17 at 04:06
  • @TorstenSchoeneberg because although I asked for an explicit extension such that segments in $\mathbb{R}$ are connected, that was because I didn't know precisely what I was looking for. Thanks to your answer I have a clearer idea. And I don't actually need it to be meaningful for transcendentals because the Collatz Conjecture probably always converges in finite steps so it's probably sufficient that it's explicit for rationals and that its extension to transcendental exists. – it's a hire car baby Dec 19 '17 at 07:56