[1.3.6] 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, let \(a \in A\) and \(b \in B\) be arbitrary elements. We know that by the definition of the least upper bound we must have,

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

Let’s move \(b\) to the other side of the inequality.

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

\(a\) here is an arbitrary element in \(A\) so \(\sup (A + B) - b\) must be an upper bound for \(A\). But we know that \(\sup A\) is the least upper bound. Therefore, by the defintion of the least upper bound, we must have

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

We can swap \(\sup A\) and \(b\) now to see that

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

With the same argument we used previously here we can see that \(\sup (A+B) - \sup A\) is an upper bound for \(B\) since \(b\) is arbitrary. Therefore,

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

Re-arranging the terms gets us

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

as required. This concludes the second part of the proof. Therefore, \(\sup A + \sup B \leq \sup (A + B)\) as required. \(\blacksquare\)

Side Note

I struggled with this proof because I was trying to figure out in my head how to transition from \(a + b \leq \sup A + \sup B\) to \(\sup (A+B) \leq \sup A + \sup B\) by trying to swap \(a + b\) with \(\sup (A+B)\). But that not how it works! The first statement establishes that \(\sup A + \sup B\) is some upper bound on \(A+B\). But because the least upper bound is always less than or equal than ANY any other bound then we can just conclude immediately that \(\sup(A+B) \leq \sup A + \sup B\).

References