TITLE: If $X=\{x_n\}, Y = \{y_n\}$ be bounded sequences of real numbers. Then, if $x_n \leq y_n~\forall~n$, then $\lim \inf (x_n) \leq \lim \inf(y_n)$
QUESTION [6 upvotes]: If $X=\{x_n\}, Y = \{y_n\}$ be bounded sequences of real numbers. Then, if $x_n \leq y_n~\forall~n$, then show that $\lim \inf (x_n) \leq \lim \inf(y_n)$ and $\lim \sup (x_n) \leq \lim \sup (y_n)$
Attempt: $\lim \inf (x_n)$ represents the least limit point in $X$ and $\lim \inf(y_n)$ represents the least limit point of $Y$.
Given that,  $~\forall~m, x_m \leq y_m, $  I am still not able to see any compulsion why the least limit point of $x_n$ should be less than the least limit point of $y_n$.
That is, for each index $m, x_m \leq y_m$. But, that doesn't place any restriction on the least limit point of $x_n$ becoming greater than the least limit point of $y_n$?
Where could I be going wrong? 
Thank you for your help.

REPLY [3 votes]: You have $x_n' = \inf_{k \ge n} x_k \le y_n$, and $x_n' \uparrow \liminf_n x_n$.
Now let $\epsilon>0$, then for $n$ greater than some $N$, we have $x_n' \ge \liminf_n x_n -\epsilon$, and so $y_n \ge \liminf_n x_n -\epsilon$. Noting that the right hand side is a constant, we have $\liminf_n y_n \ge \liminf_n x_n -\epsilon$. Since $\epsilon>0$ was arbitrary, we have the desired result.
Since $\limsup_n a_n = - \liminf_n (-a_n)$, we have the other direction.