Space-filling curve with distance locality Is there a space-filling curve $C$ that has the property that, if $C$ passes through $p_1=(x_1,y_1)$
at a distance $d_1$ along the curve, and through $p_2$ at $d_2$, then if
$|p_1 - p_2| \le a$, then $|d_1 - d_2| \le b$, for some constants $a$ and $b$?
In other words, any two points of the plane within distance $a$ are separated
by at most $b$ along $C$.  Call this property distance locality.
So I am asking whether a curve exists mapping $\mathbb{R}$ to $\mathbb{R}^2$ with distance locality.
Although I doubt the answer differs,
permit me also to ask the same question for $\mathbb{Q}^2$,
and for $\mathbb{Z}^2$. 
I have little experience with the properties of the known space-filling curves.  Those better schooled on this topic can likely answer these questions easily.  Thanks!
Addendum.
I noticed a paper just released today which is focused on "locality properties"
of 3D space-filling curves: "An inventory of three-dimensional Hilbert space-filling curves," arXiv:1109.2323v1 [cs.CG]. The author explores several different
locality measures that have been considered in the literature, and cites a wealth
of references.

Let $f:{\bf Z}\to{\bf Z}^2$ be one-one and onto. Suppose that for all pairs of adjacent lattices points $p$ and $q$ there exist integers $b$, $m$, and $n$ such that $f(m)=p$, $f(n)=q$, and $|m-n|\le b$. Suppose without loss of generality that $f(0)=(0,0)$. For any positive integer $d$, if $p$ is a lattice point $d$ steps away from the origin, then we must have $f(n)=p$ for some $n$ with $|n|\le bd$. But the number of lattice points $d$ steps away from the origin grows as the square of $d$, while the numbers of integers $n$ with $|n|\le bd$ grows linearly with $d$, contradiction. So there isn't a map satisfying "distance locality", even for distance $a=1$.  
