Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

marked as duplicate by Asaf Karagila, William, Thomas, Martin Sleziak, J. M. Oct 2 '12 at 14:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
What are $a$ and $b$ here? Sets, sequences, individual numbers? –  Tanner Swett Sep 22 '12 at 3:16
    
What setting are you working in? Subsets of $\Bbb R$? Subsets of an arbitrary partially ordered set with guaranteed existence of least upper bounds? –  Cameron Buie Sep 22 '12 at 3:17

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer
    
For the case of the two family indexed set, i don't quite get what you mean by the stronger statement is not true in general. Do you mean $$\sup_{i\in I}(a_i+b_i)\le\sup_{i\in I}a_i+\sup_{i\in I}b_i\;.$$, if you does, could you briefly explain why it is not true? As far as i know, that inequality should always holds –  abc Sep 23 '12 at 11:18
    
@abc: No, by the stronger statement I meant the statement that the two sides are equal, which was the claim that you were actually using in your argument. –  Brian M. Scott Sep 23 '12 at 19:50
    
o i see. Thanks –  abc Sep 27 '12 at 17:27
    
the second interpretation is what I assumed. I added a short answer for that interpretation, which I hope is clear. –  robjohn Oct 2 '12 at 0:55

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.

share|improve this answer

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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.