This is just alternative proof that uses lemma 1.3.8 to the proof in here. The first part of the proof is exactly the same but proving the second part is different.

Given sets A and B, define the set \(A + B = \{a + b : a \in A \text{ and } b \in B\}\). Prove that \(\sup A + B = \sup A + \sup B \)

Definitions of bounds: This.

Proof

Let \(A\) and \(B\) be sets and let \(A + B\) be the set defined above. We want to prove that \(\sup A+B = \sup A + \sup B\). To do this, we will prove:

  • \(\sup (A+B) \leq \sup A + \sup B\).
  • \(\sup (A+B) \geq \sup A + \sup B\).

From these two inequalities we will then conclude that \(\sup A+B = \sup A + \sup B\).
To prove that \(\sup A+B \leq \sup A + \sup B\), let \(a \in A\) Then we must have,

$$ \begin{align*} a &\leq \sup A. \end{align*} $$

Similarly, Let \(b \in B\). Then we must have,

$$ \begin{align*} b &\leq \sup B. \end{align*} $$

Adding both inequalities will result in

$$ \begin{align*} a + b &\leq \sup A + \sup B. \end{align*} $$

Since \(a\) and \(b\) were arbitrary, this means that \(\sup A + \sup B\) is an upper bound for \(A + B\). But we know the least upper bound for \(A+B\) is \(\sup A + B\). Then by the definition of the least upper bound, we must have

$$ \begin{align*} \sup (A+B) &\leq \sup A + \sup B. \end{align*} $$

as required. To prove condition two, define an arbitrary \(\epsilon > 0\). By lemma 1.3.8 we know that there exists some element \(x \in A\) such that,

$$ \begin{align*} x > \sup A - \epsilon/2. \end{align*} $$

(Side note: the lemma states that \(\sup A\) is a least upper bound iff for any \(\epsilon > 0\), there exists some element in \(A\) such that \(a > \sup A - \epsilon\) so really any tiny epsilon taken away from \(\sup A\) will render it not a least upper bound anymore)
Similarly, there exists an element \(y \in B\) such that,

$$ \begin{align*} y > \sup B - \epsilon/2. \end{align*} $$

Adding both inequalities we see,

$$ \begin{align*} x + y > \sup A + \sup B - \epsilon. \end{align*} $$

So we have two thing. From part 1 of the proof, we know that \(\sup A + \sup B\) is an upper bound for \(A+B\). Second, we proved above that for any \(\epsilon > 0\), there exists some elements \(x \in A\) and \(y \in B\) such that \(x + y > \sup A + \sup B - \epsilon\). Therefore, by lemma 1.3.8, \(\sup A + \sup B\) is a least upper bound for \(A + B\) and so we must have,

$$ \begin{align*} \sup A + \sup B \leq \sup (A + B). \end{align*} $$

as required. Finally, since we showed that \(\sup (A+B) \leq \sup A + \sup B.\) and that \(\sup A + \sup B \leq \sup (A + B)\), then we know that \(\sup (A+B) = \sup A + \sup B\) as we wanted to show. \(\blacksquare\)

References