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Find a sequence $\langle X_n\rangle$ such that

$$L_{X_n}=\left\{\frac{n+1}{n}:n\in\mathbb N\right\}\cup \{1\}.$$

Where $L_{X_n} = \{ p\in \mathbb{R} : \text{There exists a subsequence } \langle X_{n_k}\rangle \text{ of } \langle X_n\rangle \text{ such that} \lim X_{n_k} = p\}$.

Justify your answer.

Since given $L_{X_n}$ is an infinite set, it was difficult for me to find a sequence in which I could select infinite ordered unending set of real numbers ( Subsequences) in such a way that they converge to the given real numbers in $L_{X_n}$.

Can anyone please help me in order to solve this?

*An edit was made to define $L_{X_n}$

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  • $\begingroup$ Look up the proof of the Bolzano-Weirstrass theorem, may help $\endgroup$
    – Wolfy
    May 6 '15 at 14:13
  • $\begingroup$ Bolzano-Weirstrass theorem says that any bounded sequence has a convergent subsequence ,But here we have a find a particular sequence in such a way that we could select infinite number of subsequences which converges to those numbers in L. $\endgroup$
    – Razor1692
    May 6 '15 at 14:17
  • $\begingroup$ ok, but what is L? seems that we need to find some sub-sequence that converges to some $\mathbb{R}$ $\endgroup$
    – Wolfy
    May 6 '15 at 14:19
  • $\begingroup$ Check the problem.L is defined :) $\endgroup$
    – Razor1692
    May 6 '15 at 14:29
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Pick a set of sequences $a_n(m)$ that have limits $(n+1)/n$, or $a_0$ has limit $1$. Then interleave them. Remember the proof that the rationals are countable.

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For example $$\left( 2, 2,\frac{3}{2}, 2,\frac{3}{2} , \frac{4}{3},2,\frac{3}{2} , \frac{4}{3} ,\frac{5}{4} ,...\right)$$

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  • $\begingroup$ Thank you.I thought about constructing a constant sequence but never got the idea that i could do it this way.Thanks :) $\endgroup$
    – Razor1692
    May 6 '15 at 15:05
  • $\begingroup$ A problem!! we have to remove 1 from the set in order to define the sequence like this right? if we define the sequence like you said ,there wont be a subsequence which will converge to 1?? So with 1 ,is it impossible to construct a sequence? $\endgroup$
    – Razor1692
    May 6 '15 at 15:07
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The easiest solution to this problem is to construct a sequence such that each element in your set $L_{X_n}$ (aside from 1) appears infinitely many times. That way there is a constant subsequence, which will obviously converge to the thing that you want it to.

Here is a sequence that ought to have the needed property. I'm hiding it with a spoilers tag in case you want to try and work it out on your own.

$\left\{ \frac{2}{1},\frac{2}{1},\frac{3}{2},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\frac{7}{6},\frac{2}{1},\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5},\frac{7}{6},\frac{8}{7} \dots \right\}$

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  • $\begingroup$ Thank you.I thought about constructing a constant sequence but never got the idea that i could do it this way.Thanks :) $\endgroup$
    – Razor1692
    May 6 '15 at 15:05
  • $\begingroup$ A problem!! we have to remove 1 from the set in order to define the sequence like this right? if we define the sequence like you said ,there wont be a subsequence which will converge to 1?? So with 1 ,is it impossible to construct a sequence? $\endgroup$
    – Razor1692
    May 6 '15 at 15:07
  • $\begingroup$ No it's not impossible for many reasons. First, you could just insert 1 at the beginning of each of the descending chains, but also 1 is the limit of a subsequence already. Consider the subsequence {2,3/2,4/3,5/4,6/5,7/6,...} that converges to 1. $\endgroup$ May 6 '15 at 16:36
  • $\begingroup$ Okay.So does there exist a sequence such that L - {1} .? Which means if we consider the set L without 1 $\endgroup$
    – Razor1692
    May 6 '15 at 17:49
  • $\begingroup$ No. L will have the property that it's closed. i.e. If there is a sequence of elements in L, let's call it {y_n} and y_n converges to something, say y, then y must be in L. So, since 1 is the limit of n+1/n, if L contains infinitely many of the n+1/n it must also contain 1. $\endgroup$ May 6 '15 at 20:52

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