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Let A and B be two nonempty bounded sets of real numbers. Let

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

Show that C is a bounded set and that,

$$ sup (C) = sup (A) + sup (B) $$ and $$ inf (C) = inf (A) + inf (B)$$

What I know so far -

Bounded sets means that they have an upper bound and a lower bound. I don't know if they are complete, i.e., have a LUB or supremum.

Yeah, that is basically the knowledge I am trying to build off of. I need a kick in the right direction.

Thank you in advance!

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  • $\begingroup$ Do you want an intuitive answer or a proof? Or an intuitive proof? :) $\endgroup$ – Anthony P Dec 4 '16 at 22:58
  • $\begingroup$ Why not both? :) $\endgroup$ – ChocolateAndMath Dec 4 '16 at 22:59
  • $\begingroup$ Any bounded set in $\mathbb{R}$ has a supremum in $\mathbb{R}$. You don't need the set to be complete, just use the fact that $\mathbb{R}$ is! $\endgroup$ – Scott Burns Dec 5 '16 at 0:04
  • $\begingroup$ Right, and it is already given to us in the question that the sets are bounded. Right. But, isn't this the same as the thoery of completeness axiom "any nonempty subset of R that is bounded above has a least upper bound. In other words, the Completeness Axiom guarantees that, for any nonempty set of real numbers S that is bounded above, a sup exists" from - math.illinois.edu/~ajh/347.summer14/completeness.pdf $\endgroup$ – ChocolateAndMath Dec 5 '16 at 0:10
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Clearly for all $a \in A$ and $b \in B$ we have $a + b \le \sup A + \sup B$, which means that $C$ is bounded and $\sup C \le \sup A + \sup B$.

To see the other inequality, let $a \in A$ and $b \in B$ be arbitrary. Then $a = (a + b) - b \le \sup C - b$ holds, which means that for any $b$ the value $\sup C - b$ is an upper bound of $A$, i.e. $\sup A \le \sup C - b$. This is equivalent to $b \le \sup C - \sup A$, which implies $\sup B \le \sup C - \sup A$. This is equivalent to $\sup A + \sup B \le \sup C$, which is what we wanted to prove.

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If A is a bounded set, then

$$ \exists \quad a \in A \quad | \quad a \geq a_i \quad \forall \quad a_i\in A $$

i.e. $\sup(A) = a$

Likewise for B.

$$ \exists \quad b \in B \quad | \quad b \geq b_i \quad \forall \quad b_i\in B $$

where I'm using $i$ as a natural index for the elements in $A$ and $B$ respectively. Somewhere there ought to be the condition that $i \in \mathbb{N}$.

Now you have the set $C=A+B$. Suppose there exists an element of $C$. All of the things in $C$ are made up of the things in $A$ and $B$. The largest thing in $C$ must be the sum of the two largest things in $A$ and $B$.

$$ \forall \quad c_i \in C \qquad \sup(A)+ \sup(B) \geq c_i $$

So then $\sup(A) + \sup(B) = \sup(C)$. Of course you can show this rigorously, but maybe what I have done here is enough for you to do that using the definition of supremum? Then continue with the definition of infimum.

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  • $\begingroup$ Gotcha! I think I can take it from here :D $\endgroup$ – ChocolateAndMath Dec 4 '16 at 23:09
  • $\begingroup$ Feel free to post your complete proof as an answer so we can review it :) $\endgroup$ – Anthony P Dec 4 '16 at 23:10
  • $\begingroup$ So, from your very coherent and clear logic, I undid your efforts and got this - Let A and B be two nonempty bounded sets of real numbers. Then, $$\exists a \in A, $$ in which $$a \ge a_i, \for all A$$ and $$i \in \N$$ $\endgroup$ – ChocolateAndMath Dec 4 '16 at 23:39
  • $\begingroup$ So, sup(A) = a. |||ly, $$\exists b \in B$$, in which $$b \ge b_i, \forall b_i \in B$$ and $$i \in N$$ So, sup(B) = b $\endgroup$ – ChocolateAndMath Dec 4 '16 at 23:46
  • $\begingroup$ We are given the set, $$ C = \{ a + b: a \in A, b \in B \}. $$ Suppose, $$\exists c \in C, in which c \ge c_i, \for all c_i \in C. $$ and $$ C = a + b (given) So, $$\foralll c_i \in C, $$ sup (A) + sup (B) $$ \ge c_i$$ $$ sup (A) + sup (B) $$ = c$$ implies that, sup (A) + sup (B) = sup (C) We can |||ly, prove for infimum using its definition. $\endgroup$ – ChocolateAndMath Dec 4 '16 at 23:48

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