Characterization of convergence with liminf and limsup

Question

I am reading a text which claims the following:

Assume that \begin{align} \tag{1} \liminf_{n \rightarrow \infty} x_n \leq a < b \leq \limsup_{n \rightarrow \infty} x_n. \end{align} Then for any $\varepsilon > (b-a)/2$ we have \begin{align} \tag{2} \sup_{m \leq k < \infty} |x_k - x_m| \geq \varepsilon \end{align} for all $m \geq 0$.

Now I wonder about the choice of $\varepsilon$. Why should (2) hold for large $\varepsilon$? It seems to me that (2) holds precisely for all $\varepsilon \in [0,b-a]$. Is this correct?

Answer 1

Agree with the response given. The problem is probably with the hypothesis of the question because given its current state if we claim that

Suppose that $\lim_{x\rightarrow \infty} \sup x_n= b > a= \lim_{x\rightarrow\infty} inf x_n$.
Then $\sup_{0\leq m<k\leq \infty}|x_k-x_m|=b-a$.

Thus if we pick $\epsilon=b>b-a$, then the claim will not work anymore. I am not too sure why they made such a choice for the $\epsilon$ given that $\epsilon $ can take any value.