proving a inequality about sup 
Possible Duplicate:
How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$ 

I want to prove that  $\sup\{a+b\}\le\sup{a}+\sup{b}$ and my approach is that I claim $\sup a+ \sup b= \sup\{\sup a + \sup b\}$ and since $\sup a +\sup b \ge a+b$ the inequality is proved. Is my approach correct?

Perhaps this is what you are looking for. Consider
$$
\sup_{x\in X}(a(x)+b(x))=\color{#C00000}{\sup_{{x\in X\atop y\in X}\atop x=y}(a(x)+b(y))\le\sup_{x\in X\atop y\in X}(a(x)+b(y))}=\sup_{x\in X}a(x)+\sup_{x\in X}b(x)
$$
The red inequality is true because the $\sup$ on the left is taken over a smaller set than the $\sup$ on the right. The equalities are essentially definitions.
