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?

  • $\begingroup$ 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. $\endgroup$
    – Kapil
    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, $\exists x\in S: x^*-\varepsilon<x<x^*$. Since $x$ is the limit of some subsequence of $(x_n)$, there's infinitely many $n$ such that $x^*-\varepsilon < x_n$.


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