Need help on understanding the theorem in real analysis! 
Theorem 1.3.3: Let $L \in R$ and let $x_n$ be a sequence of real
  numbers. Then $x_n \to L$ if and only if $\limsup x_n =
 \liminf x_n = L$.

Proof: First, suppose $x_n \rightarrow L.$ Thus if $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $ n \geq N$ implies $|x_n-L| < \epsilon/2$, which implies $s_n = \sup(T_n) \leq L + \epsilon/2$ and $i_n = \inf(T_n) \geq L-\epsilon/2$. 
Thus
$$L-\frac{\epsilon}{2} \leq i_n \leq s_n \leq L +\frac{\epsilon}{2}$$
which implies that $|s_n-L| \leq \frac{\epsilon}{2} \lt \epsilon$ and $|i_n-L| \leq \frac{\epsilon}{2}<\epsilon$, for all $\epsilon > 0$. Thus $s_n \rightarrow L $ = lim sup $x_n$ and $i_n \rightarrow L$ = lim inf $x_n$.
I only included the first part of the proof. I boxed the part where I was confused. Why did they use  inequalities 
$s_n = \sup(T_n) \leq L + \epsilon/2$ and $i_n = \inf(T_n) \geq L-\epsilon/2$ instead of 
$s_n = \sup(T_n) \lt L + \epsilon/2$ and $i_n = \inf(T_n) \gt L-\epsilon/2$. 

Maybe you are thinking that the strict inequality is valid because $|x_n-L|<\epsilon/2 $ is true, but note that this inequality is true for any element $x_n$ in the sequence beyond $N$, but the supremum or infimum of such elements in $T_n$ may not be any of the elements in the tail of the sequence, so you need to write it as a weak inequality. Also note that this inequality does not affect the proof anyway. 
