This question stems from another one, but presents my concern in a more specific way.
There is a theorem saying that:
$\lim \sup x_n = \sup{z_k}$ where $z_k$ is a set of limit points for the sequence $x_n$.
All sets and sequences are real.
Limit point for a sequence is a point such that in any neighborhood of it there is a point from the sequence other than the original point itself.
$$\lim \sup x_n = \lim_{n \to \infty} \sup_{m \ge n}\{x_m\}=\inf_{n} \sup_{m \ge n}\{x_m\}$$
Now we consider a sequence $1,1,2,\frac{1}{2},...,n,\frac{1}{n},...$, for it $\limsup x_n=\infty$, but there is only one limit point $0$, thus $\sup z_k=0$. So $\limsup x_n \ne \sup z_k$. Is the theorem wrong?
In your example, we have $\limsup x_n=\infty$, so we need to work on an extension of the real line including $\infty$. The usual way of defining a topology on the extended real line is to define rays $(a,\infty]$ as the basic open neighborhoods of $\infty$. In this case, your sequence also has $\infty$ as a limit point, given by the sequence $z_k=k$, and we have $\limsup x_n=\infty=\sup z_k$. The theorem is correct if we read it this way.