([KRD10, 5.6.H]) Show that a continuous function \(f:\mathbb R\to\mathbb R\) cannot take on every real value exactly twice.
Solution
1. Suppose, toward a contradiction, that every real value is taken exactly twice.
2. Pick two preimages of one value.Proof: fix any \(x_0\); let \(a < b\) be the two points with \(f(a) = f(b) = f(x_0)\) (exactly two exist by assumption).
3. On \([a,b]\), \(f\) attains a maximum \(M\) at some \(x^* \in [a,b]\) and a minimum \(m\) at \(x_* \in [a,b]\) (extreme value theorem).
4. \(M\) is attained only at \(a\) and \(b\) (or \(f \equiv f(a)\), impossible).Proof: if \(x^* \in (a,b)\) and \(M > f(a)\), then for every \(y \in (f(a), M)\), the intermediate value theorem gives a preimage of \(y\) in \((a, x^*)\) and another in \((x^*, b)\) — at least two preimages of \(y\) inside \((a,b)\), so exactly two, both in \((a,b)\), for uncountably many \(y\).Now consider \(y = f(a)\): by IVT on \([x^*, b]\) (since \(f(x^*) = M > f(a) = f(b)\)), there is \(z \in (x^*, b)\) with \(f(z) = f(a)\), giving three preimages \(a, z, b\) of \(f(a)\) — contradiction.Hence either \(M = f(a)\) or \(x^* \in \{a, b\}\); in either case \(f \le f(a)\) on \([a,b]\), with \(f(a)\) attained at \(a, b\) and nowhere in \((a,b)\) (else a third preimage).
5. By symmetry, the minimum is \(m = f(a)\) as well.Proof: applying 4 to \(-f\) (or repeating the argument with min), \(f \ge f(a)\) on \([a,b]\).
6. Contradiction.Proof: 4 gives \(f \le f(a)\) on \([a,b]\) and 5 gives \(f \ge f(a)\) on \([a,b]\); hence \(f \equiv f(a)\) on \([a,b]\), so the value \(f(a)\) is taken uncountably many times — contradicting “exactly twice”.
7. Q.E.D.