$X_n\leq Y_n$ implies $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$ 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 $\lim\inf X_n = \inf E$.


I could have sworn I asked a question like this but I can't find it in my question list. I recall a proof of this question and it went similar to this: 
If $u>\lim sup(y_n)$, then there can be only a finite number of $n\in\mathbb{N}$ such that $u<y_n$. Since $x_n\le y_n$ then $\lim sup(x_n)\le u$ and so $\lim sup (x_n)\le \lim sup(y_n)$. 
