Regarding property of limit supremum of a bounded sequence $(x_{n})$

Let $$X = (x_{n})$$ be a bounded sequence. Limit superior of $$X$$ is $$x^* = limsup(x_{n})$$ then $$\forall \epsilon > 0$$ there exists atmost finite number of $$n \in \Bbb{N}$$ such that $$x^* + \epsilon < x_{n}$$ but infinite number of $$n$$, such that $$x^* - \epsilon < x_{n}$$. Seems puzzling to me. How do I prove this?

• One characterisation of $\limsup$ is that it is the largest number which is the limit of some sub-sequence of the given sequence. This should clarify the above statement. Apr 30 '19 at 3:01

Let $$S = \left\{x \in \left\{ x_n \right \} \middle | \text{a subsequence of {x_n} converges to x} \right\}$$ An equivalent definition of $$\operatorname{lim sup}$$ is $$\operatorname{lim sup} \left(x_n\right) = x^*= \operatorname{sup}S$$.
Let $$\varepsilon > 0$$.
Suppose $$x^* + \varepsilon < x_m$$ for infinitely many $$m$$. Then $$(x_m)$$ is a bounded sequence and thus has a convergent subsequence. This sequence converges to $$x> x^*$$, contradicting the fact that $$x^*$$ is an upper bound.
Since $$x^*$$ is a least upper bound, $$\exists x\in S: x^*-\varepsilon. Since $$x$$ is the limit of some subsequence of $$(x_n)$$, there's infinitely many $$n$$ such that $$x^*-\varepsilon < x_n$$.