2.3.19 - Let $\{x_n\}$ be a bounded sequence and $\epsilon>0$ is given. Prove that there exists an $M$ such that for all $k\geq M$, $x_k-\limsup_{n\rightarrow\infty} x_n<\epsilon$ and $\liminf_{n\rightarrow\infty}-x_k<\epsilon$.

Here's my attempt: Suppose $\{x_n\}$ is a bounded sequence and $\epsilon>0$ is given.

Now let $s_n=\sup\{x_k:k\geq n\}$ then $x_n<s_n$ for all $n\in\mathbb{N}$. The sequence $\{s_n\}$ is monotonically decreasing and is bounded below so it converges to $\inf\{s_n:n\in\mathbb{N}\}=\limsup_{n\to\infty} x_n=\ell$. Thus $s_n\geq\ell$ for all $n\in\mathbb{N}$ and $s_k-\ell<\epsilon$ for all $k\geq M$ for some $M\in\mathbb{N}$ but then $x_k-\ell\leq s_n-\ell<\epsilon$ for all $k\geq M$.

Let $a_n=\inf\{x_k:k\geq n\}$ then $x_n\geq a_n$ for all $n\in\mathbb{N}$. Now the sequence $\{a_n\}$ is monotonically increasing and bounded above so it converges to $\sup\{a_n:n\in\mathbb{N}\}=\liminf_{n\to\infty} x_n=b$ and so $a_n\leq b$ for all $n\in\mathbb{N}$. Thus $b-x_k<\epsilon$ for all $k\geq M$.

Apparently, I didn't find an $M$. How do I do that? Any help would be greatly appreciated.

  • $\begingroup$ @C-RAM how did you get the post to accept the part from "Let a_n...". I couldn't do it. $\endgroup$ Oct 24, 2022 at 3:18
  • $\begingroup$ If the claim was true, wouldn't it prove that every bounded sequence is convergent? $\endgroup$
    – gist076923
    Oct 24, 2022 at 3:22
  • $\begingroup$ Because iirc the limit exists iff liminf = limsup $\endgroup$
    – gist076923
    Oct 24, 2022 at 3:23
  • $\begingroup$ @gst076923 Read the statement more carefully, it is only making a statement about one side of each limit. $\endgroup$
    – copper.hat
    Oct 24, 2022 at 3:33
  • $\begingroup$ Ah okay I see now $\endgroup$
    – gist076923
    Oct 24, 2022 at 3:35

1 Answer 1


You are required to prove the existence of an $M$ such that, for all $k≥M$ a certain condition holds. You have found such an $M$ in your attempt. You are not required to find a value for $M$.
You have solved the problem. Good work mate!

  • 1
    $\begingroup$ This should be a comment. $\endgroup$
    – amWhy
    Oct 24, 2022 at 16:21
  • $\begingroup$ Yes, that was my intention. But I am just starting out here and I don't have enough reputation points to comment :( $\endgroup$
    – Dhiraj Rao
    Oct 24, 2022 at 16:28
  • $\begingroup$ That's what I thought but my instructor told me I didn't find a M and I got the problem wrong. Are you sure I didn't miss something or maybe I need to make another clarifyin statement? I'm now very confused as to what he wants and he won't give any more feedback. $\endgroup$ Oct 27, 2022 at 8:23

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