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There are two sets $A$ and $B$ which are bound and are not empty. Now we'll define another set as: $A + B = \{a + b | a \in A, b \in B\}.$

I need to prove that $\sup(A + B) = \sup A + \sup B.$

I don't know if the way in which I proved it is enough or is missing something. This is my proof:

I know that for every $a \in A, b \in B$: $\sup A \geq a$ and $\sup B \geq b.$ Therefore $\sup A + sup B \geq a + b.$ I'll take a $c \in A+B$ and by $A+B$ definition: $c = a + b$. Then, it is safe to say that $c = a + b \leq \sup A + \sup B$ which gives: $c \leq \sup A + \sup B.$

I know that $c \leq \sup A + \sup B$ is true for every general $c \in A+B$, so the supremum of $A + B$ is: $\sup(A + B) = \sup A + \sup B.$

Is this a way to prove it or is it not enough? Thanks.

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3 Answers 3

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I think you actually mean set instead of group. And no, that is not a correct proof. E.g., $x \le 1 \forall x \in \{0\}$ but $\sup \{0\} = 0$.

Hint:

As @Gibbs pointed out in the comments, you showed $\sup (A + B) \le \sup A + \sup B$. Now for the equality you need to show $\sup (A + B) \ge \sup A + \sup B$ also holds. Take $\epsilon > 0$,

$$ a \in A : |a - \sup A| < \frac \epsilon 2 $$ $$ b \in B : |b - \sup B| < \frac \epsilon 2 $$

which exist by characterization of supremum. Can you take it from here?

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  • $\begingroup$ So now I get sup A - ϵ/2 < a and sup B - ϵ/2 < b. I add those and get: sup A + sup B - ϵ < a + b ≤ sup(A + B) which then gives: sup A + sup B - ϵ < sup(A + B). How do I now get rid of the ϵ? Can I simply say that since we "picked" it I'll assume that even without it sup A + sup B ≤ sup(A + B)? $\endgroup$ Nov 8, 2017 at 21:30
  • $\begingroup$ Yes, that's a pretty standard argument (since it holds for every $\epsilon > 0$). For some more rigor suppose to the contrary $\sup (A + B) < \sup A + \sup B$. Then $\exists \epsilon > 0 : \sup (A + B) < \sup A + \sup B - \epsilon < \sup A + \sup B$ which contradicts what we just showed. $\endgroup$
    – cronos2
    Nov 8, 2017 at 21:57
  • $\begingroup$ I understood your proof using contradiction why sup(A+B)<supA+supB is impossible, but I still don't fully understand why removing -ϵ is valid, since it makes the small side of the inequality larger. It doesn't make sense to me. Won't removing -ϵ at some cases cause the small side of the inequality to become bigger than the right side, leaving us with sup(A+B)<supA+supB? $\endgroup$ Nov 8, 2017 at 22:25
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    $\begingroup$ Nope, that's actually what I proved in my previous comment. You can remove all the $\sup$ gibberish and play around with, e.g., $x - \epsilon < y \forall \epsilon > 0 \implies x \le y$ to convince yourself (adapting the former proof to this more general case, of course) $\endgroup$
    – cronos2
    Nov 8, 2017 at 22:28
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As Gibbs pointed out, till now we have:

$sup(A+B)\leq supA + supB$

We'll proceed by Contradiction.

Suppose, $sup(A+B)=M, M<supA+supB$

Using Denseness of $\mathbb{Q}$,

$\exists r\in \mathbb{Q}: M<r<supA+supB$

$ supA+supB \geq sup(A+B)\geq a+b$ $\forall a\in A, \forall b\in B$

$supA+supB\geq a+b>r\implies r\in A+B$

But, $r>M \implies M \neq sup(A+B) \implies sup(A+B)\geq supA+supB$

But, $supA+supB\geq sup(A+B) \implies sup(A+B)=supA+supB$

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That's not really enough as you didn't really relate $c$ to $\sup (A+ B)$.

First the boring part: showing $A+B$ is non-empty and bounded above. That's boring but needs to be done but is trivially done by noting if $a\in A, b\in B$, $A$ bounded by $c$ and $B$ bounded by $d$ then $a+b \in A+B$ and $A+B$ bounded by $c + d$.

Actually that last clause needs a little exposition. If $f \in A+B$ there are $a \in A; b\in B$ so that $f = a + b$. As $a\le \sup A$ and $b\le \sup B$ we know $f = a+b \le \sup A + \sup B$ so $A+B$ is bounded above by $\sup A + \sup B$ so $\sup (A+B) \le \sup A + \sup B$.

We need to show that if $g < \sup A + \sup B$ then $g$ can not be an upper bound of $A + B$. Or in other words there exists an $f = a+b; a\in A; b\in B$ so that $g < f=a+b \le \sup A + \sup B$. Intuitively we want $g= a' + b'$ where $a'<\sup A$ and $b' < \sup B$ so that, there will be $a,b,$ so that $a' < a\le \sup A$ and $b' < b \le \sup B$. But can we split $g$ like that?

Let $\epsilon = (\sup A + \sup B)$ then $g = [\sup A - \frac {\epsilon}2] + [\sup B -\frac {\epsilon}2]$. By def of $\sup$ we know there is an $a\in A; [\sup A - \frac {\epsilon}2] < a \le \sup A$ and a $b\in B;[\sup B - \frac {\epsilon}2] < b \le \sup B$. So $g < a + b \le \sup A + \sup B$ and, of course, $a+b \in A+B$.

So $g$ is not an upper bound of $\sup A + \sup B$. So $\sup A + \sup B = \sup (A+B)$.

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