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Let $S=\{\frac{m}{2^n}| n\in\mathbb{N}, m\in\mathbb{Z}\}$, is $S$ a dense set on $\mathbb{R}$?

Martin Sleziak
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Riemann
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    Magic words are "dyadic rationals." – Arturo Magidin May 23 '12 at 03:29
  • Didn't a similar question with 10 as the denominator surface a couple of days ago? – copper.hat May 23 '12 at 03:49
  • @copper.hat: I believe that that one asked for a little more, namely, an explicit sequence converging to any given real. – Brian M. Scott May 23 '12 at 03:53
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    @BrianM.Scott: You're right: http://math.stackexchange.com/questions/147614/is-the-given-set-dense-in-mathbbr/147618#147618 – copper.hat May 23 '12 at 03:56
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    Choose $x \in \mathbb{R}$, and let $x_n = \frac{\lfloor 2^n x \rfloor}{2^n}$. Then $x_n \in S$, $\forall n$, and $|x-x_n| < \frac{1}{2^n}$. Hence $x_n \rightarrow x$, so $S$ is dense. – Riemann May 23 '12 at 04:09
  • More generally, an additive subgroup $G$ of $\mathbb R$ is either discrete or dense and this is decided by whether $\inf { x \in G : x>0 }>0$. – lhf Aug 17 '12 at 20:56

3 Answers3

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Hint: Every real number has a binary expansion.

Quinn Culver
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Yes, is it, given open interval $(a,b)$ (suppose $a$ and $b$ positives) you can find $n\in\mathbb{N}$ such that $1/2^n<|b-a|$. Then consider the set:

$$X=\{k\in \mathbb{N}; k/2^n > b\}$$

This is a subset of $\mathbb{N}$, for well ordering principe $X$ has a least element $k_0$ then is enought taking $(k_0-1)/2^n\in(a,b)$.

The same is if $a$, $b$ or both are negatives (because $(a,b)$ is bounded).

Gaston Burrull
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This set looks really close to the rationals. Maybe you could use the density of the rationals and see if you can put an element of this set between any two rationals and go from there.

toypajme
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