Let $(x_n)_{n\in\mathbb N}$ be a sequence of real numbers. If it is bounded, then the Bolzano–Weierstrass theorem tells us that the set of accumulation points in $\mathbb R$ is non-empty. Furthermore, the set has a minimum and a maximum - the limit inferior and the limit superior.
Now let us consider the more general case where the sequence is not necessarely bounded and let $\widehat{\mathbb R}$ be the extended real number line (a totally ordered set). We can separately define what it means that $-\infty$ and $\infty$ are accumulation points. Am I right in assuming that the set of of accumulation points in $\widehat{\mathbb R}$ is always non-empty and has a minimum as well as a maximum (the limit inferior and the limit superior)?
If a sequence is not bounded above, then $\infty$ is an accumulation point, and moreover this is precisely the limit superior of the sequence. Likewise, if a sequence is not bounded from below, $-\infty$ is an accumulation point of the sequence, and also equals the limit inferior.
Otherwise, as you know, a bounded sequence has a nonempty set of accumulation points with maximum its limit superior and minimum its limit inferior.
In any case, the set of accumulation points is nonempty with minimum and maximum (possibly in the extended sense) the limit inferior and limit superior.