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


For the definitions of an upper bound and the least upper bound of a set. See 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 \(a\) be an arbitrary element in \(A\). Then we must have,

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

Similarly, Let \(b\) be an arbitrary element 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 \(\sup A + B\) and by the definition of the least upper bound \(\sup A + B\) must be less than or equal to any other bound on \(A + B\). Therefore we can write,

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

And this concludes the first part of the proof. 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 \(A\) by the definition of an upper bound. But we know that \(\sup A\) is the least upper bound and the least upper bound is less than or equal to any other bound. Therefore, we can write,

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

We can move 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 and we know that the least upper bound is less than or equal to any other bound. 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. \(\blacksquare\)

Additional Notes:

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 then we can just conclude immediately that \(\sup(A+B) \leq \sup A + \sup B\).

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