# Is there any connection between these two limsup definitions?

The limsup for a set sequence $\{E_k\}$ where $E_k, k=1,2,3...$ are sets is defined as $\lim \sup {E_k} = \bigcap\limits_{j = 1}^\infty {\bigcup\limits_{k = j}^\infty {{E_k}} }$.

The limsup for a number sequence $\{a_n\}$ where $a_n, n=1,2,3,...$ are numbers is defined as $\lim \sup {a_n} = \mathop {\lim }\limits_{j \to \infty } \mathop {\sup }\limits_{k \ge j} {a_k}$, that is, the upper bound of sequence ${a_k},{a_{k + 1}},{a_{k + 2}},...$ when $k\rightarrow\infty$.

If I understand correctly, both limsup seem different. The first limsup is a set that contains elements that belong to infinite many $E_k$s, and the second limsup is just a number.

Since they are both denoted as "limsup", I suppose there is some underlying connection between them. So is there such a connection and can you help explain it? Thank you.

A possible connection (there are others): Define the set $E_k=\{x\in \mathbb{R} \mid x \leq a_k\}$. Then, $\cup_{k=j}^\infty E_k = \{ x \in\mathbb{R}\mid \exists k \geq j, x\leq a_k\}$, and $$\lim\sup E_k=\bigcap_{j=1}^\infty \bigcup_{k=j}^\infty E_k = \{ x \in\mathbb{R}\mid \forall j\geq 1, \exists k \geq j, x\leq a_k\}$$ and if I'm not mistaken the supremum of the RHS is the $\lim\sup$ of the sequence $(a_n)_n$.
Remember that $\sup$ exists for a totally ordered set. Suppose, $$A_1\subseteq A_2\subseteq\ldots$$
is a totally ordered sequence of sets with $\subseteq$ be the order relation.
$$\sup_nA_n=\bigcup_{n=1}^\infty A_n.$$