TITLE: Show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N} \} $ is dense in $\mathbb {R}$
QUESTION [7 upvotes]: Possible Duplicate:
Density of a Set on $\mathbb{R}$? 

I have to show that show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N}\} $ is dense in $\mathbb {R}$.
A set A is dense in $\mathbb {R}$ if $\overline A=\mathbb {R}$.
But also $Y$ is a subset of $X$, we say that $Y$ is dense in $X$, if for every $x\in X$ , there is $y \in Y$  that is arbitary close to $x$.
So ,I have to prove that for every $x \in \mathbb {R}$ ,there is a number $\frac{m}{2^n}$ arbitrarily close to $x$.So $\forall \epsilon,x ,\exists y$ such that $|y-x|<\epsilon$. 
I got a little stuck at this point...Could anyone give me a hint?Thanks a lot!

REPLY [2 votes]: Do not worry about the downvote. It is an attack on my answers.
I'll sketch the proof for you. If $x$ is a real number then one can find $a_1$ and $a_2 \in A$ such that,
$$ a_1 < x < a_2 $$
Choosing $a_1 = \frac{m-1}{2^n} $ and  $a_1 = \frac{m+1}{2^n} $ gives
$$ \frac{m-1}{2^n} < x < \frac{m+1}{2^n} \Rightarrow - \frac{1}{2^n}<x-\frac{m}{2^n} <  \frac{1}{2^n} \Rightarrow  \left|x-\frac{m}{2^n}\right|<\frac{1}{2^n}=\epsilon $$