Let $S=\{\frac{m}{2^n}| n\in\mathbb{N}, m\in\mathbb{Z}\}$, is $S$ a dense set on $\mathbb{R}$?
Density of the set $S=\{m/2^n| n\in\mathbb{N}, m\in\mathbb{Z}\}$ on $\mathbb{R}$?
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real-analysis
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0More 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. – 2012-08-17
3 Answers
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Hint: Every real number has a binary expansion.
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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).
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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.