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It is easy to construct a metric space $E_d$ such that all points of $E_d$ are at mutually integral distance and such that there is a map $\varphi$ from $E_d$ into the $d$-dimensional Euclidean space such that $\varphi$ preserves distances up to a bounded error and such that $\varphi(E_d)$ is uniformly dense (sufficiently large spheres centered at points of $\varphi(E_d)$ cover the Euclidean space): Take $\mathbb Z^d$ and define the distance between two distinct points $a,b\in\mathbb Z^d$ as the integer closest to $\lVert a-b\rVert+3$ (the $+3$ is probably not optimal).

Is there a more natural metric space with the same properties? Is there such a space which is optimal in some sense?

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  • $\begingroup$ What you have is already nice, and it is a good start toward more advanced methods and finer results; this belongs to a future chapter on the join theory of metric spaces and diophantine approximations. $\endgroup$
    – Wlod AA
    Commented Sep 26, 2022 at 22:28
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    $\begingroup$ Why do you need to add 3? $\endgroup$ Commented Sep 26, 2022 at 23:02
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    $\begingroup$ TeX note: \parallel spaces poorly for norms; compare $\parallel a - b\parallel + 3$ \parallel a - b\parallel + 3 to $\lVert a - b\rVert + 3$ \lVert a - b\rVert + 3. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Sep 27, 2022 at 0:39

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(Please, do not "correct" my notation).


Let's go a bit further than in OP's q. -- let the said map (injection) into the Euclidean space be actually a bijection (or simply the identity).

Let $\ E^n\ $ be an n-dimensional space in $\ \mathbb R^n\ $ with metric $\ \rho.\ $

Every symmetric map $\ \delta: X\times X\to\mathbb R\ $ such that $\ \delta(x\ x)=0\ $ and $\ 1\le \delta(x\ y)\le 2\ $ for arbitrary $\ x\ y\in X,\ $ is a metric; in particular when $\ X:=\mathbb R^n.\ $ Let's define

$$ \forall_{x\ne y;\ x\ y\,\in\,\mathbb R^n}\quad \delta(x\ y)\ := \ 1 + \lceil \rho(x\ y)\rceil - \rho(x\ y) $$ and $\ \delta(x\ x)=0.\ $ Then $\ \delta\ $ is a metric in $\ \mathbb R^n\ $ hence so is $\ d\ :=\ \rho+\delta.\ $ We see that the identity map $\ \mathbb I_{\mathbb R^n}\ $ satisfies an inequality that improves on the inequality from OP's post:

$$ \rho\,\ \le\,\ d\ <\ \rho+2 $$

Of course, $\ d\ $ assumes integer distances only.

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    $\begingroup$ I know you are quite particular about your notation, but you had exactly one occurrence of $R^n$, which seemed very likely to be a mistake. I changed it to $\mathbb R^n$ for consistency with the other occurrences (and also ‘OT’ to ‘OP’). I hope that was all right. $\endgroup$
    – LSpice
    Commented Sep 27, 2022 at 0:40
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    $\begingroup$ @LSpice, thank you, that was helpful. #### In general, I mean the arrogance and rudeness of a number of MO participants and of MO's administration when they allow themselves to interfere with my notation free of comas, with the $\ (a;b)\ (a;b]\ [a;x)\ [a;b]\ $ notation for the intervals, etc. And there is much more to it. Then, Ironically, they -- in particular MO administration -- show their power and "teach" me politeness -- all this is so ugly. (MO administration has suspended me twice -- as the old saying says: ** give a man power, and he'll make the angels cry**. $\endgroup$
    – Wlod AA
    Commented Sep 27, 2022 at 2:38
  • $\begingroup$ I'm glad your notation is free of comas, but haven't made up my mind about the commas. [But I think I don't like $(x\ x)$]. $\endgroup$ Commented Sep 27, 2022 at 3:35
  • $\begingroup$ @GerryMyerson, sorry for my funny spelling. ###### One may not to like something but another thing is to impose on others. While MO is just a pathetic tiny nothing on the scale of the human world, the distinction between not to like vs Impose is FUNDAMENTAL. In this human world, it is the imposing that is the source of the whole human tragedy. $\endgroup$
    – Wlod AA
    Commented Sep 27, 2022 at 4:11
  • $\begingroup$ I would say that death is the source of the whole human tragedy, but I fear I'd be going off-topic if I did. $\endgroup$ Commented Sep 27, 2022 at 5:47

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