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I came across the following problem that says:

The set of limit points of the sequence $1,\dfrac12,\dfrac14,\dfrac34,\dfrac18,\dfrac38,\dfrac58,\dfrac78,\dfrac1{16},\dfrac3{16},\dfrac5{16},\dfrac7{16},\dfrac9{16},\ldots$ is which of the following?
(a) $[0,1]$
(b) $(0,1]$
(c) the set of all rational numbers in $[0,1]$
(d) the set of all rational numbers in $[0,1]$ and of the form $m/2^n$ where $m$ and $n$
are integers.

Please help. Thanks in advance for your time.

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3  
So, what do you know about limit points of sequences? Can you prove, for example, that $0$ is in the set? That will rule out (b), and maybe help you see how to do the problem. –  Gerry Myerson Dec 31 '12 at 14:58
    
It says to choose the best option. –  learner Dec 31 '12 at 14:58
    
@GerryMyerson thanks a lot sir for your feedback. –  learner Dec 31 '12 at 15:01

3 Answers 3

up vote 6 down vote accepted

The given sequence is a listing of all the rational numbers in $[0,1]$ of the form $\frac{m}{2^n}$ for some integers $m>0$ and $n\geq 0$. This set of rationals turns out to be dense in $[0,1]$, so if by "limit points" you mean "points having infinitely many terms of the sequence in every neighborhood aroud them", then the answer is $[0,1]$.

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Let $x\in[0,1]$. For any $\epsilon>0$, there is a positive integer $n$ such that $\frac{1}{2^n}<\epsilon$. Then there is a positive integer $m$ such that $x\neq\frac{m}{2^n}\in (x-\epsilon,x+\epsilon)\cap[0,1]$. But $\frac{m}{2^n}$ is an element of the given sequence. Hence the set of limit points is $[0,1]$.

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1  
Saw the anger in the chatroom, so +1. –  Tim Dec 31 '12 at 15:40

Hint: Consider the first 8 terms (reordered): $$\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}, \frac{8}{8}$$

This pattern continues as the denominator grows throughout the sequence.

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Thanks a lot Bryce for your feedback. –  learner Dec 31 '12 at 15:05

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