lim sup, lim inf, and inequalities for $a_n \le b_n$ Suppose we have two sequences ${a_n}$ and ${b_n}$, which satisfies $ a_n \le b_n$ for $n=1,2,3,\ldots$. Do we have the following inequalities to be true?
$$\limsup_{n \to \infty} a_n \le \limsup_{n \to \infty} b_n $$
$$ \liminf_{n \to \infty} a_n \le \liminf_{n \to \infty} b_n$$
In the case when  $a_n$ and $b_n$ has a limit, the equations above are clearly true, since $a_n \le b_n $ implies $\lim_{n \to \infty} a_n \le \lim_{n \to \infty} b_n $. I'm just wondering in a general case when the limits of the two sequences don't necessarily exist, are these two inequalities still true?
 A: They are both true. For each $n$, $a_k \le b_k \le \sup_{j \ge n} b_j$ for all $k \ge n$. Thus $\sup_{k \ge n} a_k \le \sup_{k \ge n} b_k$ for all $n$. Taking limits, $\limsup_{n \to \infty} a_n \le \limsup_{n\to \infty} b_n$. A similar argument produces the second inequality. For each $n$, $\inf_{j \ge n} a_j \le a_k \le b_k$ for all $k \ge n$. Thus $\inf_{k \ge n} a_k \le \inf_{k \ge n} b_k$ for all $n$. Taking limits results in $\liminf_{n\to \infty} a_n \le \liminf_{n\to \infty} b_n$.
A: Both inequalities are true. To see this, recall that $$\lim\sup\{c_n\}=\lim_{m \to \infty}\left(\sup\{c_k\mid k \geq m\}\right)$$
Fixing $A_m = \sup\{a_k \mid k \geq m\}$ and $B_m$ accordingly, it's clear that $A_m \leq B_m$ for all $m$. 
If that weren't true, we could put $A_m = B_m + \varepsilon$. Since we can always find an element within $\varepsilon$ of the supremum, we find an element $a_i$, $i \geq m$ which is greater than $B_m = \sup\{b_k \mid k \geq m\}$ showing in particular that $a_i > b_i$, a contradiction.
So your statement follows from the fact that for two sequences $\{c_i\}$ and $\{d_i\}$ satisfying $c_i \leq d_i$, $$\lim_{n \to \infty} c_i \leq \lim_{n \to \infty}d_i$$ which you can find in any calculus textbook. In this particular case the two sequences are $A_i$ and $B_i$, showing the statement for $\sup$, and the proof for $\inf$ amounts to replacing $\sup$ with $\inf$ in the above.
