A question in my book I am studying says to let $s_n$ and $t_n$ be sequences and suppose there exists $N_0$ such that $s_n \le t_n$ for all $n \gt N_0$. Show $\lim \inf s_n \le \lim \inf t_n$ and $\lim \sup s_n \le \lim \sup t_n$.
The hint for the problem in the back of my book says to let $u_N = \inf \{s_n : n \gt N\}$ and $w_N = \inf \{t_n : n \gt N\}$. Then it says that $u_N$ and $w_N$ are increasing sequences and $u_N \le w_N$ for all $N \gt N_0$.
What I don't understand is how we can assume that $u_N$ and $w_N$ are increasing sequences. From my understanding, since we don't explicitly know what $s_n$ and $t_n$, we cannot know if these are increasing or decreasing right? Also would they still be increasing if it was defined instead as $u_N = \sup\{s_n : n \gt N\}$?
Recall that the inf of a set of real numbers is the greatest lower bound. This means that for any set $S$ we have:
(1) $x\ge \inf S$ for every $x\in S$, and
(2) if $b$ is a lower bound for $S$ (i.e., $x\ge b$ for every $x\in S$), then $\inf S\ge b$.
From (1) and (2) we can prove:
Fact 1: If set $A$ is contained in set $B$, then $\inf A\ge\inf B$.
Proof: For every $x\in A$, we have $x\in B$, since $A$ is a subset of $B$. So by (1), $x\ge \inf B$ for every $x\in A$. Therefore $\inf B$ is a lower bound for set $A$. So by (2), $\inf A\ge\inf B$.
What does this have to do with your question? Note that the set $$\{s_n:n>N+1\}$$ is contained in the set $$ \{s_n:n>N\}$$ and therefore $$u_{N+1}:=\inf\{s_n:n>N+1\}\ge\inf\{s_n:n>N\}=:u_N.$$
Therefore the $\{u_N\}$ form an increasing sequence (actually, non-decreasing). And if we were taking sups instead of infs, the sequences of sups would be decreasing (actually, non-increasing), because of
Fact 2: If set $A$ is contained in set $B$, then $\sup A\le\sup B$.
which you should be able to prove analogously to Fact 1. The intuition behind facts 1 and 2 is: When a set loses elements, the inf can only get bigger, while the sup can only get smaller.