$X_n\leq Y_n$ implies $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$

HINT: Suppose that $X_n\le Y_n$ for each $n\in\Bbb N$, and suppose, to get a contradiction, that $\liminf_nY_n<\liminf_nX_n$. Then there is a subsequence $\langle Y_{n_k}:k\in\Bbb N\rangle$ that converges to some $y<\liminf_nX_n$. Show that either $\langle X_{n_k}:k\in\Bbb N\rangle\to-\infty$, or $\langle X_{n_k}:k\in\Bbb N\rangle$ has a subsequence that converges to some $x\le y$; both are impossible, since $y<\liminf_nX_n$.

The other inequality can be proved in a very similar way.

The definition of limsup/liminf for a sequence $ \{a_n: n \geq 1 \} $ is given by

$$ \liminf_{ n \to \infty} a_n = \sup_{n \geq 1} \inf_{ k \geq n } a_k \text{ and } \limsup_{n \to \infty} = \inf_{n \geq 1} \sup_{ k \geq n } a_k . $$

Let us show the inequality for $\liminf $, the other being similar. Define, for a sequence $ \{a_n \} $ and $ n \geq 1$, set $$ I_n (a) = \inf \{ a_k : k \geq n \}. $$ Clearly, $ I_n (a) \leq I_{n+1} (a) $ since the infimum is taken over $ \{ k : k \geq n \} $ in the first case which contains the set $ \{ k : k \geq n+1 \}$ over which the infimum is taken in the second case. Thus, $ I_n (a) $ is non-decreasing and hence the limit of $ I_n (a) $ exists and the supremum is the limit (which need not be bounded). Thus, we have, $$ \liminf_{ n \to \infty} a_n = \sup_{n \geq 1} \inf_{ k \geq n } a_k = \lim_{ n \to \infty} I_n (a) . $$ Now, coming to the problem, let $ X_n \leq Y_n $ for all $ n \geq 1$, we claim $$ I_n (X) \leq I_n (Y) \text{ for all } n \geq 1 . $$ Once we show this, the inequality follows from the above relation about $ \liminf $ and limit of the sequence of $ I_n (X) $ and $ I_n (Y) $ respectively.

To prove the claim: fix $ n \geq 1 $ and take any $ \epsilon > 0 $. Then $ I_n (Y) + \epsilon > I_n (Y) $ and hence there is a $ k (\geq n ) $ such that $ Y_k < I_n (Y) + \epsilon $. Thus, we have $ X_k \leq Y_k < I_n (Y) + \epsilon $. Thus, we have $ I_n (X) \leq I_n (Y) + \epsilon $. Since $ \epsilon > 0 $ is arbitrary, we have $ I_n (X) \leq I_n (Y) $ and the claim is proved.