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*}<br /> X &amp;=\{x \in \mathbb{R} : a_n &gt; x \text{ for infinitely many } n \} \\<br /> Y &amp;=\{y \in \mathbb{R} : b_n &gt; y \text{ for infinitely many } n \} \\<br /> Z &amp;=\{z \in \mathbb{R} : a_n + b_n &gt; z \text{ for infinitely many } n\}<br /> \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 &gt; z - a_n for infinitely many n. Since these a_n are bounded, they have a least upper bound x. Hence b_n &gt; 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.