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?
 A: Your notation isn’t very clear: there are at least two different things that you might be trying to prove. In one you have sets $A$ and $B$ of real numbers, and you want to show that $$\sup_{a\in A\atop{b\in B}}(a+b)\le\sup_{a\in A}a+\sup_{b\in B}b\;.$$
If so, your argument begins with the claim that
$$\sup_{a\in A\atop{b\in B}}(a+b)=\sup\left\{\sup_{a\in A}a+\sup_{b\in B}b\right\}\;.$$
This is simply assuming something stronger than what you want to prove: $\sup_{a\in A}a+\sup_{b\in B}b$ is a single number, so taking its supremum does nothing, and 
$$\sup\left\{\sup_{a\in A}a+\sup_{b\in B}b\right\}=\sup_{a\in A}a+\sup_{b\in B}b\;.$$
Thus, you’re trying to prove that 
$$\sup_{a\in A\atop{b\in B}}(a+b)\le\sup_{a\in A}a+\sup_{b\in B}b$$
by assuming the stronger statement that 
$$\sup_{a\in A\atop{b\in B}}(a+b)=\sup_{a\in A}a+\sup_{b\in B}b\;,$$
which is clearly illegitimate.
Just suppose that 
$$\sup_{a\in A\atop{b\in B}}(a+b)\not\le\sup_{a\in A}a+\sup_{b\in B}b\;,$$
i.e., that $$\sup_{a\in A\atop{b\in B}}(a+b)>\sup_{a\in A}a+\sup_{b\in B}b\;,$$
and work for a contradiction; there is one very close at hand.

The other possibility is that you have two families indexed by the same set, $\{a_i:i\in I\}$ and $\{b_i:i\in I\}$, and you want to prove that
$$\sup_{i\in I}(a_i+b_i)\le\sup_{i\in I}a_i+\sup_{i\in I}b_i\;.$$
Here again you’re assuming something stronger than what you’re trying to prove, and in this case the stronger statement is false in general. The approach that you should be using is the same as in the other interpretation: suppose that 
$$\sup_{i\in I}(a_i+b_i)\not\le\sup_{i\in I}a_i+\sup_{i\in I}b_i\;,$$
and get an easy contradiction.
And if this interpretation is the right one, you should try to find an example in which 
$$\sup_{i\in I}(a_i+b_i)<\sup_{i\in I}a_i+\sup_{i\in I}b_i\;;$$
they do exist.
A: 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.
A: It is better to think about what the inequality says than just doing formal manipulations.  On the left-hand side you have less freedom than on the rigt-hand side, this is easier to see if you write the argument of $a$ and $b$ explicitely, as such:
$$ \sup_x\{a(x)+b(x)\} \le \sup_x a(x) + \sup_x b(x)
$$
On the LHS, you must vary $x$ simultaneously in $a$ and in $b$, on the RHS you can vary $x$ separately in $a$ and in $b$. In that way, every value you can get on the LHS can be matched on thje RHS, but not in the opposite direction. That makes the inequality obvious.
