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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!

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    @QuinCulver Thanks!I was trying to find it.I knew that there could be this question,how I didn´t find it,I posted.Thanks for letting me know!2012-08-17

4 Answers 4

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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}

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    I would have started with let \epsilon>0 be given, then there is an $n\in\Bbb N$ such that 1/2^n<\epsilon. Then consider the set $\{k/2^n: k\in\Bbb Z\}$. Then there is a least $k$ such that x. This $k$ is your $m+1$. But great proof otherwise; I don't know why anyone would downvote it.2014-06-16
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$\frac{\lfloor 2^nx\rfloor}{2^n}\leqslant x\lt \frac{\lfloor 2^nx\rfloor+1}{2^n}$

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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.

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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.