Since ε is arbitrarily small, do the inequalities hold?

[yucheng]

##$If$b \leq x_n \leq c$for all but a finite number of n, show that$b \leq \operatorname{lim inf}_{n \to \infty} x_n$and$\operatorname{lim sup}_{n \to \infty} x_n \leq c_n$(Buck, Advanced Calculus, Section 1.6, Exercise 24) Let$\beta =\operatorname{lim inf}_{n \to \infty} x_n$and$\alpha = \operatorname{lim sup}_{n \to \infty} x_n$. Let$\varepsilon$be any number greater than 0. Since$\beta$and$\alpha$are limit points, there exists subsequences of integers$\{n_k\}$and$\{n_i\}$, both infinite, such that$|x_{n_k}-\beta| \leq \varepsilon$and$|x_{n_i}-\alpha| \leq \varepsilon$Then,$-\varepsilon \geq x_{n_k}-\beta \leq \varepsilon$and$-\varepsilon \geq x_{n_i}-\alpha \leq \varepsilon$. From this, we get b \leq x_{n_k} \leq \beta + \varepsilon and \alpha - \varepsilon \leq x_{n_i} \leq c Since$\varepsilon$is arbitrarily small, is it true that the inequalities above become$b \leq \beta$and$\alpha \leq c##

 

[mfb]

Since $\beta$ and $\alpha$ and limit points, $|x_n-\beta| \leq \varepsilon$ and $|x_n-\alpha| \leq \varepsilon$

Did you mean "are"? For which n is that supposed to hold? You didn't specify. What is $\varepsilon$?
In general alpha and beta can be different, so they cannot both be the limit of the sequence.

Focus on one side, the other one is completely analogous.

 

[yucheng]

Did you mean "are"? For which n is that supposed to hold? You didn't specify. What is $\varepsilon$?
In general alpha and beta can be different, so they cannot both be the limit of the sequence.

Focus on one side, the other one is completely analogous.

I apologize for my inaccurate language. Let me fix my post.

 

[wrobel]

I prefer the following definition
$$\limsup x_n=\lim_{n\to\infty}\sup_{k\ge n}\{x_k\}.$$
With this the assertion is clear

 

[yucheng]

I prefer the following definition
$$\limsup x_n=\lim_{n\to\infty}\sup_{k\ge n}\{x_k\}.$$
With this the assertion is clear

Would you clarify which assertion? And what do you mean by $\operatorname{sup_{k \geq n}}$ (specifically $k \geq n$)? Thanks!

 

[wrobel]

Would you clarify which assertion?

the assertion you are supposed to prove:
if $a\le x_n\le b\quad \forall n\in\mathbb{N}$ then $\limsup x_n\le b,\quad \liminf x_n\ge a$

And what do you mean by (specifically )?

$$\sup_{k\ge n}\{x_k\}=\sup\{x_n,x_{n+1},\ldots\}$$