Solve Supremum Problem: X, Y and Z Sets

[e(ho0n3]

Homework Statement
Let $a_1,a_2,\ldots$ and $b_1,b_2,\ldots$ be bounded sequences of real numbers. Define the sets X, Y and Z as follows:

$$\begin{align*} X &=\{x \in \mathbb{R} : a_n > x \text{ for infinitely many } n \} \\ Y &=\{y \in \mathbb{R} : b_n > y \text{ for infinitely many } n \} \\ Z &=\{z \in \mathbb{R} : a_n + b_n > z \text{ for infinitely many } n\} \end{align*}$$

Show that $\sup Z \le \sup X + \sup Y$ and that equality holds if one of the sequences converges.

The attempt at a solution
Let X+Y = {x + y : x in X and y in Y}. I know that sup (X+Y) = sup X + sup Y. If I can show that Z is a subset of X+Y, then sup Z ≤ sup (X+Y) and the inequality follows.

To this end, let z belong to Z. Then $b_n > z - a_n$ for infinitely many n. Since these $a_n$ are bounded, they have a least upper bound x. Hence $b_n > z - a_n \ge z - x$. Let y = z-x and notice that y belongs to Y. Unfortunately, x does not belong to X and there seems to be no fix to this problem.

What else can I do?

 

[e(ho0n3]

By the way, sup X, sup Y, and sup Z are the definitions of $\displaystyle \limsup_{n \to \infty}a_n$, $\displaystyle \limsup_{n \to \infty}b_n$, and $\displaystyle \limsup_{n \to \infty} (a_n+b_n)$, respectively.