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Consider a set of numbers in a divergent sequence $x_n$.
My deduction:
If a set has no limit point, then there does not exist any sequence in the set that converges.
Then if {$x_n$} has no limit point, then $\forall$ subsequences of $x_n$, $x_{n_k}$ diverges.
Note that the proof of the theorem

$``$Every subsequence of a sequence diverging to infinity diverges to infinity $"$

involves the assumption that $n_k > k$ which is not the case here, since $n_k$ is not necessarily increasing.

So, how do you prove there is no limit point in a $x_n$.
(PS: you may use the theoerm above after proving $n_k$ diverges)

By diverge, I mean diverge to infinity.

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closed as unclear what you're asking by user99914, Alex Provost, José Carlos Santos, Arnaud Mortier, Taroccoesbrocco Aug 8 '18 at 17:22

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    $\begingroup$ Divergent sequence, or divergent to infinity? There's a slight difference $\endgroup$ – Jakobian Aug 8 '18 at 10:14
  • $\begingroup$ Not every subsequence of a divergent sequence is divergent. $\endgroup$ – Hagen von Eitzen Aug 8 '18 at 10:14
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    $\begingroup$ You should clarify whether by "diverging" you automatically mean "diverging to infinity", because there are other kinds of non-convergence which are all usually referred to as "divergence" (e.g. oscillation). $\endgroup$ – M. Winter Aug 8 '18 at 10:20
  • $\begingroup$ 'If a set has no limit point, then there does not exist any sequence in the set that converges.' -- how about a set $\{0\} $ and a costant sequence $(0, 0, 0 \ldots ) $? The set is a singleton, so it has no limit point but a sequence is obviously convergent. $\endgroup$ – CiaPan Aug 8 '18 at 10:22
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It is not clear what you mean by a divergent sequence, but from what you have written it appears that your sequence $x_n \to \infty $. If possible let there be a limit point $a$. Consider the interval $(a-1,a+1)$. For all $n$ sufficiently large, $x_n >a+1$ so $x_n \notin (a-1,a+1)$. This contradicts the definition of limit point.

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  • $\begingroup$ But how does this prove there is no $x_n \in (a-1, a+1)$, or in general, $(a-\delta, a+\delta)$ other than $a$ itself? $\endgroup$ – Astrick Harren Aug 8 '18 at 14:23
  • $\begingroup$ If $a$ is a limit point then any interval $(a-\delta,a+\delta)$ must contain $x_n$ for infinitely many values of $n$. $\endgroup$ – Kavi Rama Murthy Aug 8 '18 at 23:25
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The statement is not necessarily true:

Let $$a_{2n}=1-\frac{1}{2n},$$ $$a_{2n+1}={2n+1}$$

This is a sequence that is divergent, yet has a limit point - namely $1$.

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  • $\begingroup$ Yes, I should clarify that by diverge I meant diverge to infinity. $\endgroup$ – Astrick Harren Aug 8 '18 at 14:25

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