Let $A+B = \{a+b | a \in A, b \in B\}$ and denote the least upper
bound of a set $X$ as $lub(X)$.Show that $lub(A+B) = lub(A) + lub(B)$.
I'm finding this problem a little confusing. I've been provided with a proof below, but it's still not really clear to me. Can anyone elucidate this for me?
Proof: since $a \leq lub(A), b \leq lub(B)$, we have $a +b \leq lub(A) + b$ and $b +lub(A) \leq lub(B) + lub(A)$, so $a+b \leq lub(B)
+ lub(A)$, so $lub(A)+lub(B)$ is an upper bound for $A+B$. Now suppose $x < lub(A)+lub(B)$. Then $\epsilon = lub(A) + lub(B) -x > 0$. Then
$\frac{\epsilon}{2} > 0$. Since $lub(A) – \frac{\epsilon}{2} <
lub(A)$, there exists $a_0 \in A$ s.t. $lub(A) – \frac{\epsilon}{2} <
a_0$. Similarly, since $lub(B) – \frac{\epsilon}{2} < lub(B)$, there
exists $b_0 \in B$ s.t. $lub(B) – \frac{\epsilon}{2} < b_0$.Then $x = lub(A)+lub(B)-\epsilon = lub(A) – \frac{\epsilon}{2} +
lub(B) – \frac{\epsilon}{2} < a_0 + b_0$. So $x$ is not an upper bound
for $A + B$. Therefore, $lub(A) + lub(B)$ is the least upper bound.
I'm confused specifically by the last step: how does our statement imply that $x$ is not an upper bound for $A+B$?
Best Answer
Look at that last line concerning $x$:
$$x = \ldots < a_{0}+b_{0}.$$
But $a_{0}$ and $b_{0}$ were chosen as members of $A$ and $B$ respectively (members within $\epsilon/2$ of $\textrm{lub}(A)$ and $\textrm{lub}(B)$). Since $a_{0}\in A$ and $b_{0}\in B$, it follows that $a_{0}+b_{0}\in A+B$.
We have now exhibited a member of $A+B$ that is greater than $x$, (namely $a_{0}+b_{0})$, so we may conclude that $x$ cannot be an upper bound for $A+B$.
Finally, since $x$ is an arbitrary number smaller than $\textrm{lub}(A)+\textrm{lub}(B)$, it follows that no upper bound for $A+B$ can be smaller than $\textrm{lub}(A)+\textrm{lub}(B)$, which in turn proves that $$\textrm{lub}(A)+\textrm{lub}(B)\leq \textrm{lub}(A+B).$$
I take it that you followed the first part of the proof, where it was established that $$\textrm{lub}(A)+\textrm{lub}(B)\geq \textrm{lub}(A+B).$$ so that covers the entire proof.