How to show if a sequence diverges to infinity then it has no limit point?

Question

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 theorem above after proving $n_k$ diverges)

By diverge, I mean diverge to infinity.

Answer 1

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.

Answer 2

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 $1$.