# How to show if a sequence diverges to infinity then it has no limit point? [closed]

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.

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

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• Divergent sequence, or divergent to infinity? There's a slight difference – Jakobian Aug 8 '18 at 10:14
• Not every subsequence of a divergent sequence is divergent. – Hagen von Eitzen Aug 8 '18 at 10:14
• 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). – M. Winter Aug 8 '18 at 10:20
• '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. – CiaPan Aug 8 '18 at 10:22

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.
• 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? – Astrick Harren Aug 8 '18 at 14:23
• If $a$ is a limit point then any interval $(a-\delta,a+\delta)$ must contain $x_n$ for infinitely many values of $n$. – Kavi Rama Murthy Aug 8 '18 at 23:25
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 $1$.