Does an uncountable Golomb ruler exist? Does there exist an uncountable set $G\subset \mathbb{R}$ such that, for $a,b,c,d \in G$, if $a-b=c-d$ then $a=c$ and $b=d$? 
 A: I think so (assuming Choice). Let $G$ be a vector space basis of $\Bbb{R}$ over $\Bbb{Q}$. Then $G$ is, among other things, linearly independent over $\Bbb{Z}$ and uncountable.
A: Well order $\Bbb R$, and proceed by transfinite induction to define $G_\alpha$.
Suppose that for all $\alpha<\beta$, $G_\alpha$ was defined. Then $G_\beta=\bigcup_{\alpha<\beta}G_\alpha\cup\{r_\gamma\}$, where $r_\gamma$ is the least real number in the well-ordering we fixed such that adding $r_\gamma$ does not violate the property of being a Golomb ruler.
It is only needed to show that if $\bigcup G_\alpha$ is countable, then such $r_\gamma$ exists. But note that $|\{a-b\mid a,b\in\bigcup G_\alpha\}|\leq|\bigcup G_\alpha|+\aleph_0$. Therefore if $\alpha<\frak c$ then there is such $r\in\Bbb R$ for which $a-r$ and $r-a$ are unique for all $a\in\bigcup G_\alpha$. So in fact the induction can continue all the way up to $\frak c$.
The true question is whether we need the axiom of choice in order to show that such thing exists. If you can show that such set cannot have a perfect subset, then by assuming that every uncountable set of reals has a perfect set (and it is consistent with the failure of the axiom of choice) you can prove that no such uncountable Golomb ruler exists. But I don't know if this sort of set cannot contain a perfect set.
A: No choice is required. It suffices to exhibit an uncountable collection of real numbers which is linearly independent over $\mathbb{Q}$, and explicit choice-free examples are given in the answers to this MO question. 
