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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..

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  • $\begingroup$ So, if there are an infinite number of natural numbers, $n$ such that $x_n > v$, is it still possible that $v$ is your limsup, or would it have to be something bigger? Now that part1 is done, since we know that there must be a finite number of $x_n>v$, but there are an infinite number of $x_n$, what does that say about part 2? (also, i think it should be for part 2 that $x_n \leq v$ as the sequence $\{1,1,1,1,1,1...\}$ has limsup=1, and there are no $x_n\lneq 1$). $\endgroup$ – JMoravitz Oct 18 '14 at 19:15
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    $\begingroup$ I suppose both parts are false. A counterexample may be $x_n = \frac{1}{n}$. $\endgroup$ – Adayah Oct 18 '14 at 19:18
  • $\begingroup$ @JMoravitz If there are an infinite number of natural numbers, n such that $xn>v, $, then : they form a sub sequence of $X$, whose supremum must be clearly $>u$ . Hence, there can not be infinite elements greater than $v$. Am I correct? $\endgroup$ – MathMan Oct 18 '14 at 19:23
  • $\begingroup$ @VHP I'm pretty sure that just implies $\geq u$, not a strict greater than. $\endgroup$ – Bruce Zheng Oct 18 '14 at 19:25
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    $\begingroup$ As Adayah points out, the statement you are trying to prove is false. I think you want to prove that for any $epsilon>0$, 1) there are only finitely many n with $v+\epsilon<x_n$ and 2) there are infinitely many n with $v-\epsilon<x_n$. $\endgroup$ – user84413 Oct 18 '14 at 22:39
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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 $(y_n)$.

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  • $\begingroup$ $1.$ Suppose that $x_n > v + \epsilon$ for infinitely many $n$. Then for any $m∈N, {x_n:n≥m}$ contains an element which is greater than $v+ϵ.$ , which means, $\lim \sup X$ would be shifted to the right of $ L + \epsilon$ which is a contradiction. Am I correct? $\endgroup$ – MathMan Oct 18 '14 at 23:53
  • $\begingroup$ $2$. Suppose, $v−ϵ<x_n$ for only finitely many $n$. Then for some $m∈N, n≥m $ why does this \implies $x_n ≤ v−ϵ$ , as we have only a finite number of elements less than $ v - \epsilon$? $\endgroup$ – MathMan Oct 18 '14 at 23:55
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    $\begingroup$ 1) That is the right idea; this implies that $y_m>v+\epsilon$ for every m, so $v=\lim y_m\ge v+\epsilon$. 2) If $v-\epsilon<x_n$ for only finitely many n, then for n large enough, say $n\ge m$, $x_n\le v-\epsilon$, so then $y_m\le v-\epsilon$. $\endgroup$ – user84413 Oct 19 '14 at 0:05
  • $\begingroup$ I think I got it. In case $1.$ if there are infinite number of elements $> v+ \epsilon$, then the limit superior will shift to the right. and in the second case, when $2$. when there are only finite number of elements $< v - \epsilon$, the limit superior would shift to the left bringing a contradiction in both cases. I hope I am correct for both. $\endgroup$ – MathMan Oct 19 '14 at 0:53
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$$L^+=\limsup x_n:=\inf _{k\ge 1}\big(\sup_{n\ge k} x_n\big)$$

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

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