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..
Best Answer
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 $(y_n)$.