Show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N} \} $ is dense in $\mathbb {R}$ 
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!
 A: $$\frac{\lfloor 2^nx\rfloor}{2^n}\leqslant x\lt \frac{\lfloor 2^nx\rfloor+1}{2^n}$$
A: In order to prove that these numbers - called the dyadic rationals, I believe - are dense in $\mathbb R$, it suffices to show that any real number is a limit of a sequence of such numbers. For a given $x \in \mathbb R$, consider the sequence $\big( \frac{\lfloor 2^n x \rfloor}{2^n} \big)$, where $\lfloor \cdot \rfloor$ is the usual "largest-integer-less-than-or-equal-to" function.
A: A thought: Given $\epsilon > 0$ and $x\in\mathbb{R}$, there is some rational, say $\frac{p}{q}$ that is within $\epsilon/2$ of $x$.  Then can you find $m,n\in\mathbb{Z}$ such that $|\frac{p}{q} - \frac{m}{2^n}| < \epsilon/2$?  
If so, you could then apply the triangle inequality.
A: 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
$$ \frac{m-1}{2^n} < x < \frac{m+1}{2^n} \Rightarrow - \frac{1}{2^n}<x-\frac{m}{2^n} <  \frac{1}{2^n} \Rightarrow  \left|x-\frac{m}{2^n}\right|<\frac{1}{2^n}=\epsilon $$
