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If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?

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    $\begingroup$ The same way you show anything about a supremum: show that the right side is an upper bound of A+B, and then show that any other upper bound is larger. $\endgroup$ Sep 13, 2010 at 17:25
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    $\begingroup$ I have a question, that , if we consider $A = (-1)^n $ & $B= 1/n$ . In this will $SUP A+B$ =$SUP A + SUP B$? $\endgroup$
    – Rkb
    May 30, 2019 at 12:18
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    $\begingroup$ @ManthaSaiGopal The Addition you did on the set is incorrect. The set should be like $A+B = \{(-1)^m + 1/n, m,n \in N \}$, and it's easy to see the supremum of this set is 2. $\endgroup$
    – rikusp2002
    Mar 5, 2022 at 5:11
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    $\begingroup$ Because $1 \in A$ and $1 \in B$. That is why 1 + 1 = 2 must exist in $A + B$ $\endgroup$
    – rikusp2002
    Mar 5, 2022 at 5:15
  • $\begingroup$ Thanks for pointing it out, you are correct $\endgroup$ Mar 7, 2022 at 16:53

5 Answers 5

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Set $x=\sup A$, $y=\sup B$. Given $a\in A, b\in B$ we have $a\le x, b\le y$ so $a+b\le x+y$ so it's an upper bound. Now take $\varepsilon > 0$ and find $a,b$ such that $a>x-\varepsilon/2, b>y-\varepsilon/2$ and you have $a+b > x+y - \varepsilon$. That suffices (since it means that every potential "smaller upper bound" $x+y-\varepsilon$ is not really an upper bound).

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  • $\begingroup$ Why is your second line, not enough? (I'm asking why the first line is there). By uniqueness of the supremum, this says x + y = sup(A+B)? $\endgroup$
    – Travis62
    Jan 11, 2021 at 19:29
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    $\begingroup$ @Travis62 (I know it's old, but somebody might see this and remain uncertain). The given proof is divided into parts: first it is proved that the set $A+B$ is upper bounded; then the supremum is found to be $\sup A+\sup B$. Both parts are necessary. $\endgroup$
    – egreg
    Jul 25, 2022 at 9:04
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Show that $\sup(A+B)$ is less than or equal to $\sup(A)+\sup(B)$ by showing that the latter is an upper bound for $A+B$. Then show that $\sup(A)+\sup(B)$ is less than or equal to $\sup(A+B)$ by showing that $\sup(A+B)$ is an upper bound for $A+\sup(B)$ and that $\sup(A+\sup(B)) = \sup(A)+\sup(B)$.

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The following proofs are based on Question 2 here. The nub is to prove two inequalities: $$\sup(A+B) \geq \sup A+ \sup B \tag{1}$$

$$\sup(A + B) \leq \sup A + \sup B \tag{2}$$


Proof of $(2)$:

By definition of $A + B$ and $\sup(A + B)$, for all $a \in A$ and $b \in B$,

$$\color{seagreen}{a + b} \color{seagreen}{\leq \sup (A + B)}.$$ Subtract $b$ from both sides:

$$a = \color{seagreen}{a + b} - b \color{seagreen}{\leq \sup (A + B)} - b.$$

Hence, if we fix $b \in B$, then $\color{seagreen}{\sup (A + B)} - b$ is an upper bound for $$\color{seagreen}{A + B} - B = A.$$

And so by definition of $\sup A$, for every $b \in B$, $$\sup A \leq \sup (A+ B) − b.$$ Rearrange the previous inequality: $\color{magenta}{b} \leq \sup(A +B) − \sup A$ for all $b \in B$.

Hence, $\sup(A +B) − \sup A$ is an upper bound for any $\color{magenta}{b}$.

By the definition of supremum, the previous inequality means: $$\color{magenta}{\sup B} \leq \sup(A + B) − \sup A \iff \sup A + \sup B \leq \sup(A + B).$$


Proof of $(1)$:

Since $\sup A$ is an upper bound for $A$, $a \leq \sup A$ for all $a \in A$. Then $b \leq \sup B$ for all $b \in B$. Hence $a + b \leq \sup A + \sup B$ for all $x \in A$ and $y \in B$. Hence, $\sup A + \sup B$ is an upper bound for $A + B$. Hence, by definition of supremum, $\sup A + \sup B \geq \sup(A + B)$.

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  • $\begingroup$ Please correct me if I'm wrong, but I'm not entirely convinced by the latter half. $\sup (A+B)$ does not have to be part of the set $A+B$, right? So I'm not sure how your last statement holds - $\sup A + \sup B$ does not have to be $\ge A+B$ $\endgroup$
    – shrgm
    Oct 4, 2017 at 1:22
  • $\begingroup$ @shreyasgm I may be wrong. How can I correct this pls? $\endgroup$
    – user53259
    Mar 21, 2018 at 6:07
  • $\begingroup$ @TuckerRapu I improved your formatting, I hope that is alright. As an aside, why did you include the link at the top? $\endgroup$
    – user279515
    Aug 13, 2018 at 18:28
  • $\begingroup$ @Brahadeesh No problem! That was thes ource. $\endgroup$
    – user53259
    Jan 12, 2019 at 3:26
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    $\begingroup$ @TuckerRapu Oh, I see. Nice answer, btw. :) $\endgroup$
    – user279515
    Jan 12, 2019 at 4:08
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Another proof for $\sup(A + B) \geq \sup A + \sup B$.


Proof when $\sup A + \sup B$ is finite:

Posit $e > 0.$ Then there exists $a \in A$ and $b \in B$ such that $a > \sup A − \frac{e}{2}$ and $b > \sup B − \frac{e}{2}$. Then $a + b \in A + B$. Ergo, $$\color{seagreen}{\sup(A + B)} \geq a + b \color{seagreen}{> \sup A + \sup B - e} \implies \color{seagreen}{ \sup(A + B) > \sup A + \sup B - e }.$$ Since $e > 0$ is arbitrary, $\sup(A + B) \geq \sup A + \sup B$

Proof when $\sup A + \sup B$ is infinite:

Since the sets here are nonempty, the suprema here are not equal to $-\infty$, so we're not in danger of encountering the undefined sum $-\infty +\infty$. If $\sup A + \sup B = + \infty$, then at least one of the suprema, say $\sup B$, equals $+\infty$. Select some $a_0 \in A$. Then $$\sup(A+B) \geq \sup(a_0 + B) = a_0 + \sup B = + \infty,$$ so $\sup(A+B) \geq \sup(A) + \sup(B)$ holds in this case. (Source)

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$$\forall ~~a\in A,~~b\in B, a\le \sup A~~ \text{and }~~b\le \sup B \implies a+b \le a\le \sup A +\sup B$$

hence $$\sup A +B\le \sup A +\sup B$$ Now we know there exist $a_n\in A$ and $b_n\in B$ such that $$\sup A=\lim_{n\to\infty} a_n\qquad and \qquad\sup B =\lim_{n\to\infty}b_n$$

then $$\sup A+\sup B =\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n =\lim_{n\to\infty} a_n+b_n \le \sup A+B$$

whence $$\sup A+\sup B = \sup A+B$$

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  • $\begingroup$ did you assume here that $\lim_{n\to\infty} a_n+b_n = \sup A+ B $ ? if so, why? $\endgroup$
    – sky1099
    Jul 24, 2022 at 14:18
  • $\begingroup$ For every n, $a_n+b_n\leq\sup A+ B$ now you can pass to the limit $\endgroup$
    – Guy Fsone
    Jul 25, 2022 at 8:56

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