| context | $A\in \text{Seq}(X)$ |
| definition | $\underset{n\to\infty}{\limsup}A_n\equiv{\bigcup_{n=1}^\infty}\left({\bigcap_{k=n}^\infty}A_k\right)$ |
We have that
$\underset{n\to\infty}{\limsup}A_n\subseteq \underset{n\to\infty}{\liminf}A_n,$
see Set limes inferior. If moreover
$\underset{n\to\infty}{\limsup}A_n=\underset{n\to\infty}{\liminf}A_n,$
then we call it
$\underset{n\to\infty}{\lim}A_n$
and say $A$ is convergent.
Wikipedia: Limit superior and limit inferior
For a more possibly more elucidating explaination, see my answer in this Math.Se thread this Math.SE thread
Parameterized set