Enumeration of $\mathbb Q^n$ such that the distance of consecutive points is $1$ in some norm of $\mathbb R^n$ For which $n \in \mathbb N$ , does there exist a bijection $f : \mathbb N \to \mathbb Q^n$ and a norm $\|\cdot\|$ on $\mathbb R^n$ such that $\|f(k+1)-f(k)\|=1, \forall k \in \mathbb N$ ?
 A: When $n= 1$, up to a multiplicative constant the absolute value is the only norm, and it is easy to see that there is no solution.
When $n\geq 2$, however, there are always solutions, if we take $||.||$ to be the "max" norm defined by $||(x_1,x_2,\ldots,x_n)||_{\infty}=\max(|x_k|; 1\leq k \leq n)$. Note that there is no solution if we take
an irrational multiple of this norm, for example $||x||=\sqrt{2}||x||_{\infty}$ (because for this norm, points in ${\mathbb Q}^n$ are always at irrational distances from each other).
A natural way to solve the problem is to start with any enumeration of ${\mathbb Q}^n$
and iterate (using the axiom of choice) the following procedure : given any point $a$ in ${\mathbb Q}^n$ not yet enumerated by $f$, find a finite path between $a$ and $b$ (where
$b$ is the last point enumerated by $f$ so far), consisting only of not-yet-enumerated-by-$f$ points, and such that the distance between two successive points
in the path is always $1$. This motivates the following definition :
Definition. Let $a,b\in{\mathbb Q}^n$. A finite path $(v_0,v_1,\ldots,v_N)$ in ${\mathbb Q}^n$ is $(N,a,b)$-nice iff $v_0=a,v_N=b$ and $||v_{k+1}-v_k||_{\infty}=1$ for $0\leq k \lt N$.
The procedure described above actually works, thanks to the following lemma :
LEMMA 1. Let $a,b\in{\mathbb Q}^n$ and let $X$ be any finite subset of ${\mathbb Q}^n$. Then, there is a finite $(N,a,b)$-nice path $(v_0,v_1,\ldots,v_N)$ such that  $v_k\not\in X$ for $1\leq k \leq N-1$.
This lemma clearly follows from another lemma :
LEMMA 2. Let $a,b\in{\mathbb Q}^n$. Then, there is a $N\in{\mathbb N}$ such that
the set of all $(N,a,b)$-nice paths (denote it by ${\cal N}(N,a,b)$) has the property that the projection $\lbrace v_k | (v_0,v_1,\ldots,v_N) \in {\cal N}(N,a,b) \rbrace$ is infinite for $1\leq k \leq N-1$.
Using concatenation and working one coordinate at a time, it suffices to show lemma 2 when
$a$ and $b$ differ by only one coordinate, the first, say: $a_1\neq b_1$. We may assume without loss that $a_1<b_1$.  
LEMMA 3. Let $a_1<b_1$ in $\mathbb Q$. Then there is an odd $N$ such that the set
${\cal M}(N,a_1,b_1)$ of all sequences $(w_0,w_1,\ldots,w_N)\in{\mathbb Q}^{N+1}$ with $w_0=a_1,w_N=b_1$, and $|w_{k+1}-w_k| \leq 1$ for $0\leq k \lt N$ has the property that the projection $\lbrace w_k | (w_0,w_1,\ldots,w_N) \in {\cal M}(N,a,b) \rbrace$ is infinite for $1\leq k \leq N-1$.
Proof of lemma 2 (from lemma 3) To go from $(w_0,w_1,\ldots,w_N)\in{\cal M}(N,a_1,b_1)$ to $(v_0,v_1,\ldots,v_N)\in{\cal N}(N,a,b)$, define
$v_k=(w_k,a_2+\frac{1+(-1)^k}{2},a_3+\frac{1+(-1)^k}{2},\ldots,a_n+\frac{1+(-1)^k}{2})$.
Proof of lemma 3. Using concatenation again, and considering the integer part of $b_1-a_1$, it suffices to show lemma 3 when $|b_1-a_1| \leq 1$. In that case, we may take $N=3$ since all the $(a_1,x,b_1)$ for $x \in (a_1,b_1)$ are in ${\cal M}(3,a_1,b_1)$ . This finishes the proof.
