# Let $x_n$ be a bounded sequence and $\epsilon$ is given. Prove $x_k-\limsup x_n<\epsilon$ and $\liminf x_n-x_k<\epsilon$.

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 $$x_n for all $$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.

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

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 $$M$$.