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Does the set {$ 1,\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{1}{16},\frac{3}{16},....$} dense in [0,1]. I am not sure about that.

I think it is not dense in [0,1] because the set contains rationals of the form $\frac{m}{2^n}$, where $m,n\in N$.

Am i correct? If not please give me a proof and also help me to gain intuition. I want intuition and hint rather than proof. Please leave the proof for me. My question in different from the previously asked question because I'm in need of intuition and the answers there doesn't give me enough satisfaction.

  • You have correctly described the set. Now see if you can understand why you can approximate any number (rational or not) using these numbers. Hint: think about bisecting intervals. – Ethan Bolker Oct 11 '17 at 13:26
  • What definition of density are you using? – Michele Maschio Oct 11 '17 at 13:37
  • I am asking this question because I'm need more sophisticated answers –  Oct 11 '17 at 13:38
  • @Michele Maschio Rudin's definition. "A subset A of a metric space M is dense if every point of M is a point of A or a limit point of A" –  Oct 11 '17 at 13:39
  • It will hit every number in the binary number system between 0 and 1. If the binary number system could not represent all real numbers arbitrary well we would have trouble! – mathreadler Oct 11 '17 at 13:55

2 Answers2

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Choose $x\in (0,1]$. If you want to approximate $x$ by these numbers in the distance of $\varepsilon > 0$, choose $N$ such that $2^{-N} < \varepsilon$. Then $$ (0,1] = (0,\frac 1 {2^N}]\,\cup\,(\frac 1 {2^N},\frac 2 {2^N}]\,\cup\,\ldots\,\cup(\frac {2^{N-1}}{2^N},1] $$ and $x$ is contained in one of those intervals.

amsmath
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THE ANSWER IS YES: In fact\

For $x\in[0,1]$ We have $$\lfloor 2^nx\rfloor \le 2^nx < \lfloor 2^nx\rfloor +1\Longleftrightarrow \frac{\lfloor 2^nx\rfloor}{2^n}\rfloor \le x \le \frac{\lfloor 2^nx\rfloor}{2^n}+\frac{1}{2^n}\\\Longleftrightarrow 0\le x-\frac{\lfloor 2^nx\rfloor}{2^n}\le \frac{1}{2^n}\to 0$$

Where $$\lfloor a\rfloor =\max\{n\in \Bbb Z: n\le a\}$$ stand for the floor of $a.$

That is $$\frac{\lfloor 2^nx\rfloor}{2^n}\to x,~~~$$ but $$\lfloor 2^nx\rfloor \in\mathbb N,~~~\forall n\in\mathbb N $$ ie

$$\frac{\lfloor 2^nx\rfloor}{2^n}\in \{ 1,\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{1}{16},\frac{3}{16},....\}~~~\forall n\in\mathbb N $$

Guy Fsone
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