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!
On
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.
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.