Characterization of convergence with liminf and limsup I am reading a text which claims the following:
Assume that
\begin{align} \tag{1}
\liminf_{n \rightarrow \infty} x_n \leq a &lt; b \leq \limsup_{n \rightarrow \infty} x_n.
\end{align}
Then for any $\varepsilon &gt; (b-a)/2$ we have
\begin{align} \tag{2}
\sup_{m \leq k &lt; \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?
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 $\lim_{x\rightarrow \infty} \sup x_n= b &gt; a= \lim_{x\rightarrow\infty} inf x_n$. 
Then $\sup_{0\leq m&lt;k\leq \infty}|x_k-x_m|=b-a$.

Thus if we pick $\epsilon=b&gt;b-a$, 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.
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 $\lim_{x\rightarrow \infty} \sup x_n= b &gt; a= \lim_{x\rightarrow\infty} inf x_n$. 
Then $\sup_{0\leq m&lt;k\leq \infty}|x_k-x_m|=b-a$.

Thus if we pick $\epsilon=b&gt;b-a$, 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.