Let $a \in \mathbb{R}$ be an accumulation value of the real sequence $(a_n)_{n \in \mathbb{N}}$ and an upper bound of the set $\{a_n | n \in \mathbb{N}\}$. Show that $$a=\sup\{ a_n | n \in \mathbb{N} \} = \limsup_{n \to \infty} a_n$$
I think I understand the equation but I do not know how to prove it in a rigorous way.
Since $a$ is an upper bound of $\{a_n\}_{n=1}^\infty$, therefore \begin{equation} L\le a\tag1\label{1} \end{equation} where $\displaystyle L:=\limsup_{n\to\infty}a_n$ (we know that limsup always exists, finitely or infinitely). Also, since $a$ is an accumulation point, for any $\varepsilon>0$, $a-\varepsilon$ has infinite number of points of the sequence above it. Hence, \begin{align} &a-\varepsilon\le L\qquad\forall\varepsilon>0\\ \implies &a\le L\tag2\label{2} \end{align} From \eqref{1} and \eqref{2}, we get the desired result.