Let $ \{x_n \}$ be an arbitrary sequence of real numbers and $A \in \mathbb R$. How prove that:
If there exists $ K \in \mathbb N $ such that for all $ n> K $, we have that $ a_n \leq A $ then $$\lim_{n\to \infty}\sup a_n\leq A.$$
If for all $K \in \mathbb N$, there exists $ n_k> K $ such that $ a_ {n_k} \leq A $, then $$\lim_{n\to\infty}\inf a_{n}\leq A.$$
I have tried but I have not achieved anything, I do not even know where to start. Someone help me with one and I make the other or some suggestion that allows me to do both items...
Definition: Let $ \{s_n \} $ be a sequence of real numbers, and in the set of numbers $ x $ (in the extended system of real numbers) such that $s_ {n_k} \to x $ for some sub-sequence $ \{s_ {n_k} \} $. This set E contains all the subsequence limits, possibly the numbers $ \pm \infty $. Let's do $$s^*=\sup E$$ $$s_*=\inf E$$ the numbers $ s^* $ and $ s_* $ are called the upper and lower limits of $ \{s_n \} $ and the notation is used $$\lim_{n\to\infty}\sup E=s^*$$ $$\lim_{n\to\infty}\inf E=s_*$$