Proof that the sup of a set is the inf of its upper bounds.

Theorem: Let A be any nonempty set of real numbers, bounded above so that \(\sup A\) exists. Define \(B = \{M\in \Bbb R: M \text{ is an upper bound for }A\}\). Then B is nonempty, bounded below by \(\sup A\), and \(\sup A = \inf B\).

Proof: Since \(\sup A\) is an upper-bound for the set A, the set B is nonempty. Trivially any element of A will be a lower-bound for B, so that B is bounded below and therefore \(\inf B\) exists.

We probably need to start with showing that \(\inf B\) is an upper bound on A. Suppose not for contradiction. Then there is an element \(a\in A\) such that \(\inf B < a\), and this in turn implies that there is an element \(b\in B\) such that \(b < a\). But this contradicts the fact that B is the set of upper bounds for A. Hence we have shown that \(\inf B\) is an upper bound on the set A.

Next we need to show that \(\inf B\) is the least of the upper bounds. In effect, though, that's really the same thing as showing that \(\inf B = \min B\), so it would be enough to show that \(\inf B\in B\). But in fact ... that's what we just showed, and therefore we shocking already have derived the fact that \(\inf B = \min B\) and by definition this is the same as \(\inf B = \sup A\).