Let \(A\) and \(B\) be nonempty subsets of \(\mathbb{R}\) with \(A\) bounded above and \(B\) bounded below. Define \(A - B = \{a - b: a \in A, b \in B\}\). Prove that \(\sup(A - B) = \sup A - \inf B\)

Proof

Since \(A\) is bounded above, then by the axiom of completeness we must have a least upper bound, \(\sup A\). Similarly, since \(B\) is bounded below, then we must have a greatest lower bound, \(\inf B\). Now, for any \(a\), we must have

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

Let \(b\) be any element in \(B\). Subtract \(b\) from each side to see that

$$ \begin{align*} a - b &\leq \sup A - b \\ &\leq \sup A - \inf B \quad \text{(because $b \geq \inf B$)}. \end{align*} $$

This establishes \(\sup A - \inf B\) as an upper bound for \(A - B\). Next, we want to show that \(\sup A - \inf B\) is the least upper bound. This means that want to show for all \(\epsilon > 0\), that there exists some element \(a - b \in A - B\) such that

$$ \begin{align*} a - b \geq \sup A - \inf B - \epsilon. \end{align*} $$

Since \(\sup A\) is the least upper bound, then we know that for any \(\epsilon > 0\), there exists some element \(a\) such that

$$ \begin{align*} a &> \sup A - \frac{\epsilon}{2} \end{align*} $$

Similarly

$$ \begin{align*} b &< \inf B + \frac{\epsilon}{2} \\ -b &> - \inf B - \frac{\epsilon}{2} \end{align*} $$

Adding both inequalities

$$ \begin{align*} a - b &> \left(\sup A - \frac{\epsilon}{2}\right) + \left(-\inf B - \frac{\epsilon}{2}\right) \\ a - b &> \sup A - \inf B - \epsilon. \end{align*} $$

As we wanted to show. \(\ \blacksquare\)

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