Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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

share|cite|improve this question
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 $\left(\sup B-\frac{d}2,\sup B\right)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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