0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
2
$\begingroup$

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 $x^*-\varepsilon < x_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.