TITLE: $X_n\leq Y_n$ implies $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$
QUESTION [14 upvotes]: Can anyone prove this question? I tried but I didn't get any I idea, so I hope someone can solve it.

Let $X_n\leq Y_n$ for each $n\in \Bbb N$. Show that $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$.

Please prove this question - thanks.
The definition I have:

Let $X_n$ be a sequence in real number and let $$E=\{x\in \Bbb R^\sharp:(X_{n_k}) \rightarrow x \text{ for some subsequence }(X_{n_k})\text{ of }(X_n)\}$$ for all $n \in \Bbb N$ and $k$ from $1$ to $\infty$. Then by definition $\lim\sup X_n = \sup E$ and $\lim\inf X_n = \inf E$.

REPLY [2 votes]: If $\lim\inf X_n=\infty$ then $X_n\rightarrow\infty$ implies $Y_n\rightarrow\infty$. Assume $\infty>\lim\inf X_n>\lim\inf ~Y_n$. For $0<\epsilon<\lim\inf X_n-\lim\inf ~Y_n$, we've $n_0\in \mathbb{N}$ such that $X_{m}>(\lim\inf X_n)-\epsilon/2$ for $m>n_0$. From the choice of $\epsilon$, there is a $n_1>n_0$ such that $(\lim\inf X_n)-\epsilon/2>Y_{n_1}$. But then we get $X_{n_1}>Y_{n_1}$, a contradiction to the assumption! So, we conclude $\lim\inf X_n\leq\lim\inf ~Y_n$. We apply this to the sequences $\lbrace-X_n\rbrace$ and $\lbrace-Y_n\rbrace$ by observing $-X_n\geq-Y_n$ to prove for the $\lim\sup$.