there are only finitely many n with $v+ϵ<x_n$ and $2)$ there are infinitely many $n$ with $v−ϵ<x_n$

Question

Prove that if $v$ is the limit superior of a bounded sequence $X$, then for any $\epsilon>0,$ $(i)$ there are only finitely many n with $v+ϵ<x_n$ and $2)$ there are infinitely many $n$ with $v−ϵ<x_n$

Attempt: Let $v$ be the limit superior of a sequence $X$.

Let $y_0,y_1,\cdots$ represent the supremum of the sequences $\{x_n,x_{n+1},x_{n+2},\cdots \}$ respectively.

Now, $Y=y_0>y_1>y_2>\cdots$ represents a monotonically decreasing bounded sequence. Hence, this sequence will converge to $\lim_{m \rightarrow \infty} y_m = \inf ~Y = v$.

Part $(i) : $Suppose there are an infinite number of natural numbers $n$ such that $x_n \geq v$

I am not sure how to proceed further. Please tell me how to proceed ahead.

Part $(ii)$ : We need to prove that there are infinite number of natural numbers $n$ such that $x_n < v$

Unfortunately, I am not sure here either.

Please guide me on how to move ahead.

Thank you for your help..

Answer 1

Hints:

1) Suppose that $x_n>v+\epsilon$ for infinitely many n. Then for any $m\in\mathbb{N}$, $\{x_n: n\ge m\}$ contains an element which is greater than $v+\epsilon$.

2) Suppose that $v-\epsilon<x_n$ for only finitely many n. Then for some $m\in\mathbb{N}$, $n\ge m\implies x_n\le v-\epsilon$.

Now show that these lead to contradictions by looking at the sequence $(y_n)$.

Answer 2

$$L^+=\limsup x_n:=\inf _{k\ge 1}\big(\sup_{n\ge k} x_n\big)$$

Let $y>L^+$, there is a $N$, s.t., for all $k\ge N$, $\sup_{n\ge k} x_n< y$, i.e., $x_n<y$ for all $n\ge N$. So for any $\alpha >L^+$, $\{n:x_n>\alpha\}$ is finite

For $y<L^+$, if we fix any $N\ge 1$, we thus have $y<\sup_{n\ge N} x_n$. So there is a $k>N$ s.t., $y<x_k$, i.e., there are infinitely many $x_n$ greater than $y$. So for any $\beta<L^+$, $\{n:x_n>\beta\}$ is infinite