Intermediate Value Theorem

1. Set up a sup.Proof: assume \(f(a) < f(b)\) and fix \(x\) with \(f(a) < x < f(b)\).Let \(E = \{t \in [a,b] : f(t) \le x\}\); then \(a \in E\) (so \(E \ne \varnothing\)) and \(E\) is bounded above by \(b\).Let \(c = \sup E\).

2. \(c \in (a,b)\).Proof: by continuity at \(a\), since \(f(a) < x\), there is \(\delta > 0\) with \(f < x\) on \([a, a+\delta)\), so \(c \ge a + \delta > a\).By continuity at \(b\), since \(f(b) > x\), there is \(\delta' > 0\) with \(f > x\) on \((b-\delta', b]\), so \(E \subseteq [a, b-\delta']\) and \(c \le b - \delta' < b\).

3. \(f(c) = x\).Proof: suppose \(f(c) < x\).By continuity at \(c\) there is \(\delta > 0\) with \(f(t) < x\) for \(t \in (c - \delta, c + \delta)\); then \(c + \delta/2 \in E\), contradicting \(c = \sup E\).Suppose \(f(c) > x\).By continuity at \(c\) there is \(\delta > 0\) with \(f(t) > x\) on \((c - \delta, c + \delta)\), so no point of \((c - \delta, c]\) lies in \(E\), making \(c - \delta\) an upper bound of \(E\) smaller than \(c\) — contradiction.Hence \(f(c) = x\).

4. Q.E.D. Proof: \(c \in (a,b)\) with \(f(c) = x\); the case \(f(a) > f(b)\) follows by applying this to \(-f\), and \(f(a) = f(b)\) cannot occur with \(f(a) < x < f(b)\).