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Let $A$ and $B$ be sets of real numbers. Define a set $A+B$ by $A+B =\{a+b|a \in A, b \in B\}$.

Show that if $A$ and $B$ are bounded sets, then $g.l.b.(A+B) = (g.l.b. A)+(g.l.b.B)$.

(The g.l.b. being the greatest lower bound.)

So I'm just really confused at how to prove this.

So far I have,

$Proof. $ Let $A,B$ be sets of real numbers and bounded sets. Let $\alpha = glbA$ and $\beta = glb B$.

$\forall x \in A,$ there exists an $x \geq \alpha$ and $\forall y \in B,$ there exists $y \geq \beta$.

Not sure where I want to go from here. Please help!

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The first thing to do would probably be to observe that $glb(A+B)\geq glb(A)+glb(B)$, since $glb(A)+glb(B)$ is easily seen to be a lower bound for $A+B$. – 1015 Feb 3 '13 at 23:13
Another method would be proof by contradiction. Assume that it's not equal. Then there are two options: the left hand side is bigger or smaller then the right hand side. Then for each case you can show that cannot be correct, then you conclude that your assumption was incorrect. – Applied mathematician Feb 4 '13 at 0:30

By contradiction:

Assume $$glb(A+B) \neq glb(A) + glb(B)$$ Then either $glb(A+B) > glb(A) + glb(B)$ or $<$.

Assume the former. Since $A$ and $B$ are bounded, $A+B$ must be bounded (can you see this?). Then, it exist some $c \in A+B$ such that:

  • $c = a' + b'$, with $a' \in A, \,\, b' \in B$
  • $c \leq x, \,\,\forall x \in A+B$. This means $c$ is the glb of $A+B$.

On the other hand, since $A$ and $B$ are bounded, there are some $\alpha$ and $\beta$ such that $\alpha = glb(A)$ and $\beta = glb(B)$

What is the relation between $a'$ and $\alpha$. and $b'$ and $\beta$? That answered, what is the relation between $a' + b'$ and $\alpha + \beta$?

Do you find anything weird, or something impossible (given the assumptions we agreed on) between the former relations and the definition of our $c$?

This is halfway done, you also need to see what happens when you assume $glb(A+B) < glb(A) + glb(B)$

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Why when you add $A +B$ does it have to be bounded? – Matt Feb 4 '13 at 5:35
A set if bounded if it has both upper and lower bounds. An upper bound for $A+B$ would be the sum of any upper bound for individual $A$ and $B$, which will give a number 'bigger' than any element in $A+B$, and these exist, because $A$ and $B$ are bounded. The same thing goes for a lower bound. Recall that, for an upper/lower bound $b$ for a set $A$, $b\in A$ does not need to hold. – Sebastialonso Feb 4 '13 at 11:47

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