Difficult to prove $\lim_ {n \to \infty} x_n - \liminf y_n = \limsup (x_n - y_n)$ I wanted to prove something here about $\liminf$ and $\limsup$, but it is not coming out, I need some help.
Let $\{x_n\}$ and $\{y_n\}$ two sequences of real numbers such that $0\leq y_n\leq x_n$ for all $n$ and $\lim_{n\to\infty}x_n$ exists. I need to prove that
$$ \lim_ {n \to \infty} x_n - \liminf y_n = \limsup (x_n - y_n). $$
Ok, I did the following:
The left side can be writen as
$$ \lim_ {n \to \infty} x_n - \liminf y_n =\lim_ {n \to \infty} x_n - \lim_ {n \to \infty}\inf\{ y_n, y_{n+1},y_{n+2},\ldots\} = $$
$$= \lim_ {n \to \infty}(x_n- \inf\{ y_n, y_{n+1},y_{n+2},\ldots\}) = \lim_ {n \to \infty}\sup \{ x_n-y_n, x_n-y_{n+1},x_n-y_{n+2},\ldots\}$$
while the right side is $$\limsup (x_n - y_n) = \lim_{n\to\infty}\sup\{x_n-y_n, x_{n+1}-y_{n+1}, x_{n+2}-y_{n+2},\ldots\}.$$
I can't see a good way to compare this sets.  
Any help is welcome, thanks!
 A: Let's define $A_n = \{x_n - y_n, x_n-y_{n+1}, x_n - y_{n+2}, \ldots\}$, $B_n = \{x_n - y_n, x_{n+1} - y_{n+1}, x_{n+2} - y_{n+2}, \ldots\}$.  Then, what we wish to prove is that $\lim\sup A_n = \lim\sup B_n$.
Let $z_n$ be an element of $A_n$ with $z_n \ge \sup A_n - \frac{1}{n}$.  We'll show that $\lim_{n \to \infty} z_n \le  \lim\sup B_n$.  Suppose $z_n = x_n - y_m$.  Then, there exists a corresponding element $w_n = x_m - y_m$ in $B_n$.  We have that $m \ge n$, so for every $\epsilon > 0$, by making $n$ sufficiently large we can guarantee
$$z_n < w_n + \epsilon \le \sup B_n + \epsilon$$
based on the fact that $x_n$ is Cauchy.  Since this is true for all $\epsilon > 0$ by making $n$ large enough, it follows that $\lim\limits_{n \to \infty} z_n \le \lim\limits_{n \to \infty} \sup B_n$.  But, $\lim_{n\to\infty}z_n = \lim_{n\to\infty}\sup A_n$, so $$\lim_{n \to \infty} \sup A_n \le \lim_{n \to \infty} \sup B_n.$$
Using the same technique, we can similarly show that $\lim\sup B_n \le \lim\sup A_n$.  It thus follows that $\lim\sup A_n = \lim\sup B_n$, which is exactly what you wanted to show.

Here's a proof I think is slightly easier.  Let's let $x = \lim_{x \to \infty} x_n$.  Then, for any $\epsilon > 0$, for $n$ sufficiently large, we have
$$x - \epsilon - y_n < x_n - y_n < x + \epsilon - y_n$$
So, taking the $\lim\sup$, 
$$x - \epsilon + \lim\sup(-y_n) \le \lim\sup (x_n - y_n) \le x + \epsilon + \lim\sup(-y_n)$$
since $\epsilon > 0$ was arbitrary, we have
$$x + \lim\sup (-y_n) \le \lim\sup (x_n - y_n) \le x + \lim\sup(-y_n)$$
So, using the well-known identity $\lim\sup (-y_n) = -\lim\inf y_n$ and observing that $x = \lim_{n\to\infty}x_n$,
$$\lim_{n\to\infty}x_n - \lim\inf y_n = \lim\sup(x_n-y_n)$$
