Just getting around to posting thoughts I had regarding this question about the additive structure of the real numbers. I was interested in which sets generate $(\mathbb{R},+)$. First, is the following argument correct? Given any set $A$ of positive Lebesgue measure, the Steinhaus theorem says that $A-A$ contains an open neighborhood of the origin. As per Arturo Magidin's answer to the original question, any such interval generates $\mathbb{R}$. Noting that $A-A$ is contained in the subgroup of $\mathbb{R}$ generated by $A$, we see that $A$ in fact generates $\mathbb{R}$.

Second, are there sets of measure zero which generate $\mathbb{R}$? I looked around a bit, but am not really sure what tools to use to approach this question.

  • 7
    $\begingroup$ $C - C = [-1,1]$, where $C$ is the Cantor set. $\endgroup$
    – martini
    Mar 22, 2012 at 15:59
  • $\begingroup$ @martini: nice ;-) $\endgroup$
    – dtldarek
    Mar 22, 2012 at 16:17
  • $\begingroup$ Wow, if I've ever seen that property of the Cantor set, I can't claim to remember it. Thanks! $\endgroup$
    – dls
    Mar 22, 2012 at 23:03

1 Answer 1


You could just take $\{1,-1\}$ and numbers from [0,1] whose representation in decimal system contains only zeros and ones. The measure of this set is $$1-\sum\frac{9^n}{10^{n+1}} = 1 - \frac{1}{10}\frac{1}{1-\frac{9}{10}} = 0.$$ To prove that this generates $\mathbb{R}$, let $x$ be the real number from [0,1] to be represented (the integer part is trivial to generate). Set $x_i$ to be the number which has ones on the places where $x$ has digit $i$ and then $x = 1x_1 + 2x_2 + \ldots + 9x_9$.

Edit: in fact this is the same as martini's comment (my set is a version of set of Cantor).


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