Suppose that $(a_{n})_{n=0}^{\infty}$ is a bounded sequence and that $L\in\textbf{R}$. Then $a_{n}\to L$ as $n\to\infty$ if and only if \begin{align*} \limsup_{n\to\infty}(a_{n}) = \liminf_{n\to\infty}(a_{n}) = L \end{align*}
Proof
I am mainly interested in the direction $(\Rightarrow)$.
The book which I am reading provides the following argument:
"Suppose that $a_{n}\to L$ as $n\to\infty$. Then if $\varepsilon > 0$ there exists $n_{0}$ s.t. $L - \varepsilon/2 < a_{n} < L + \varepsilon/2$ for $n\geq n_{0}$.
Thus $L - \varepsilon < M_{n} < L + \varepsilon$ for $n\geq n_{0}$ so that $M_{n}\to L$ as $n\to\infty$. Thus $\limsup_{n\to\infty}(a_{n}) = L$.
Similar reasoning applies to the $\liminf_{n\to\infty}(a_{n})$."
Here, the notation $M_{n}$ means $\sup\{a_{j}:(j\in\textbf{N})\wedge(j\geq n)\}$.
My question
I did not understand why $L - \varepsilon/2 < a_{n} < L + \varepsilon/2$ implies that $L - \varepsilon < M_{n} < L + \varepsilon$.
Could someone help me interpreting this?
For all $ k \geq n_0$, we have $L-\epsilon/2 < a_k < L+ \epsilon/2$.
So for all $ n \geq n_0$ $$ L-\epsilon/2 \leq \sup_{k \geq n} a_k \leq L+ \epsilon/2 $$ and thus
$$ L-\epsilon < M_n < L+ \epsilon. $$
Note how the strict inequalities become large inequalities when taking the supremum.