Based on the definition, $x$ is a limit point of $S$ if for all $\epsilon>0$ , there is a point $y \in S \backslash \{ x \}$ with $d(x,y)<\epsilon$.
I'm looking for a sequence of real numbers that has $4$ distinct limit points in the extended real line such that its superior limit is equal to $0$ and its inferior limit is equal to $-\infty$.
Can someone give me an example with an explanation please.
Let $$ a_n = \begin{cases} -n & n \equiv 0 \pmod 4 \\ -2 & n \equiv 1 \pmod 4 \\ -1 & n \equiv 2 \pmod 4 \\ 0 & n \equiv 3 \pmod 4 \end{cases} $$
Sequence begins like $$ -2,-1,0,-4,-2,-1,0,-8,-2,-1,0,-12,-2,-1,0,-16,\ldots $$ Since $-2,-1,$ and $0$ appear infinitely often, they are limit points. Obviously, $0$ is the superior limit, since it's a limit and every element is $\leq 0$, and the inferior limit is $-\infty$ since it is unbounded below.