Exercise 1.3.6

Question

Given sets $A$ and $B$, define $A+B=\{a+b: a \in A$ and $b \in B\}$. Follow these steps to prove that if $A$ and $B$ are nonempty and bounded above then $\sup (A+B)=\sup A+\sup B$
  \begin{enumerate}[label=(\alph*)] 
  \item Let $s=\sup A$ and $t=\sup B$. Show $s+t$ is an upper bound for $A+B$.
  \item Now let $u$ be an arbitrary upper bound for $A+B$, and temporarily fix $a \in A$. Show $t \leq u-a$.
  \item Finally, show $\sup (A+B)=s+t$.
  \item Construct another proof of this same fact using Lemma 1.3.8.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We have $a \le s$ and $b \le t$, adding the equations gives $a + b \le s + t$.
  \item $t \le u - a$ should be true since $u - a$ is an upper bound on $b$, meaning it is greater then or equal to the least upper bonud $t$. Formally $a + b \le u$ implies $b \le u - a$ and since $t$ is the least upper bound on $b$ we have $t \le u - a$.
  \item From (a) we know $s + t$ is an upper bound, so we must only show it is the least upper bound. \par
    Let $u = \sup(A+B)$, from (a) we have $t \le u-a$ and $s \le u-b$ adding and rearranging gives $a+b \le 2u - s - t$. since $2u - s - t$ is an upper bound on $A+B$ it is less then the least upper bound, so $u \le 2u - s -t$ implying $s + t \le u$. and since $u$ is the least upper bound $s+t$ must equal $u$. \par
    Stepping back, the key to this proof is that $a + b \le s,\forall a,b$ implying $\sup(A + B) \le s$ can be used to transition from all $a+b$ to a single value $\sup(A+B)$, avoiding the $\epsilon$-hackery I would otherwise use.
  \item Showing $s + t - \epsilon$ is not an upper bound for any $\epsilon > 0$ proves it is the least upper bound by Lemma 1.3.8. Rearranging gives $(s - \epsilon/2) + (t - \epsilon/2)$ we know there exists $a > (s - \epsilon/2)$ and $b > (t - \epsilon/2)$ therefore $a + b > s + t - \epsilon$ meaning $s+t$ cannot be made smaller, and thus is the least upper bound.
   \end{enumerate}

Exercise 1.3.6

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Exercise 1.3.6

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