Metric rigidity in $E^{n}$

Abstract:

In answer to a question rasied by L. Janos, it is shown (i) that ${E^n}$ is not a finite union of metrically rigid subsets; (ii) that if ${2^\omega } > {\kappa ^ + }$, then ${E^n}$ is not the union of $\kappa$ metrically rigid subsets, where $\kappa$ is an infinite cardinal; and (iii) that if ${2^\omega } = {\kappa ^ + }$, then ${E^1}$ is the union of $\kappa$ metrically rigid subsets, and hence that ${E^1}$ is a countable union of metrically rigid subsets iff the continuum hypothesis holds. ($A \subseteq {E^n}$ is metrically rigid iff no two distinct two-point subsets of $A$ are isometric.) Open question: assuming the continuum hypothesis, can ${E^n}$ be written as a countable union of metrically rigid subsets if $n > 1$?