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Given the sets $A, B \subseteq \mathbb R$ and $A, B \ne \varnothing,$ $A,B$ upper bound, with

$$C = \{a + b | a \in A, b \in B\},$$

I am asked to show that $\sup C = \sup A + \sup B$. I have a few questions:

  1. Since no order operation is given, but I'm using $\mathbb R,$ is it correct to assume that the order is simply the 'usual' $\le$ operation?
  2. Is the following proof correct?

$$\begin{align*} &\sup A = \max A,\sup B=\max B\tag{1}\\ &\sup C=\max C=\max A + \max B\tag{2}\\ &(1)\text{ and }(2) \implies \sup C=\sup A + \sup B \end{align*}$$

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This is false. Consider $A = B = [0,1]$. Then $C = [0,2]$. Presumably, you are asked to show that $\sup(C) = \sup(A) + \sup(B)$. – JavaMan Nov 16 '11 at 13:47
@JavaMan Ah sorry, typo. I didn't press Shift long enough. That's supposed to be a plus, not an equal. – Paul Manta Nov 16 '11 at 13:48
A few things about the problem also: Note that "\sup" will allow you to write $\sup$ instead of $sup$. Also, $\max(A) = \sup(A)$ only if the maximum exists. However, for some sets, the maximum doesn't exist (like, say $(0,1)$) so the supremum and the maximum are not always equal. – JavaMan Nov 16 '11 at 13:49
Writing \sup with a backslash not only de-italicizes it, but also causes "$\sup A$" to have proper spacing between $\sup$ and $A$, and affects the position of the subscript so that although you see $\sup_{x\in S}$ in an "inline" setting, you see $\displaystyle\sup_{x\in S}$ in a "displayed" setting. – Michael Hardy Nov 16 '11 at 16:52
up vote 3 down vote accepted

(1) Yes, the intended order is the usual one on $\mathbb{R}$.

(2) As JavaMan noted, $\max A$ need not exist: what if $A = [0,1)$, for instance? Let’s look at a concrete example in which your argument fails completely. Take $A = [0,1)$ and $B=[2,3)$; what numbers are in $C$? You should be able to convince yourself fairly readily that $C=[2,4)$. In this example none of the three sets has a maximum element, though all are bounded above.

To show that $\sup C=\sup A+\sup B$, you need to begin by showing that $\sup C$ exists, i.e., that $C$ is bounded above. Since you want $\sup A+\sup B$ to be the supremum of $C$, you might try to show first that $\sup A+\sup B$ is an upper bound for $C$, i.e., that $a+b\le \sup A + \sup B$ whenever $a\in A$ and $b\in B$. Having done this, you’d know both that $\sup C$ exists and that $\sup C \le \sup A + \sup B$.

The natural next step would be to try to show that $\sup A + \sup B \le \sup C$, so as to be able to conclude that $\sup C = \sup A + \sup B$. It’s a little hard to see right away just what it means for $\sup A + \sup B$ to be less than or equal to $\sup C$, so try assuming the opposite and getting a contradiction: what happens if $\sup C < \sup A + \sup B$? You’d get a contradiction if you could show that whenever $u<\sup A + \sup B$, there are $a\in A$ and $b\in B$ such that $a+b>u$. This is where the real work in this argument takes place.

HINT: Let $d=(\sup A + \sup B)-u$, and look at the intervals $\left(\sup A-\frac{d}2,\sup A\right)$ and $\left(\sup B-\frac{d}2,\sup B\right)$.

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