TITLE: Is there any uncountably infinite set that does not generate the reals?
QUESTION [31 upvotes]: Does there exist an uncountably infinite set $X \subseteq \mathbb R$ such that $\mathbb R \neq \left<X\right>$? I can't think of any, but I'm also having trouble trying to prove that no such subset exists.
For example: $\mathbb R$ is uncountable and obviously $\mathbb R = \left<\mathbb R\right>$. The Cantor set $C$ is uncountable, and we know that $C - C = [0, 1]$, so then since $\mathbb R = \left<[0, 1]\right>$ we know that $C$ also generates $\mathbb R$. Also the set of irrationals $\mathbb R \setminus \mathbb Q$ is uncountable, but we can generate all the rational numbers by fixing one irrational number $\alpha$ and then saying the any rational number $x$ shall be $(\alpha + x) - \alpha$, since both $\alpha + x$ and $\alpha$ are irrational.
So the examples that quickly come to mind all generate the reals. Is there a simple counterexample?

REPLY [3 votes]: generate in the sense of fields
Let $X \subseteq \mathbb R$ be a set with Hausdorff dimension zero, and furthermore all Cartesian products $X \times X \times \dots \times X$ have Hausdorff dimension zero.  Then the field $F$ generated by $X$ also has Hausdorff dimension zero (so $F$ is not all of $\mathbb R$).  You can construct Cantor sets $X$ like this, which are uncountable.  
plug
G. A. Edgar & Chris Miller, Borel subrings of the reals.  Proc. Amer. Math. Soc. 131 (2003) 1121-1129
LINK
Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.
Also: see the references there for subgroups of the reals (due to Erdös and Volkmann) with Hausdorff dimension $t$ for any $t$ with $0<t<1$.