Prove that a continuous function from \(\mathbb R\) to \(\mathbb R\) which maps open sets to open sets must be monotone.
Solution
We prove the contrapositive. Assume f is not monotone, i.e., there exist \(a < b < c\) with \(f ( a ) < f ( b )\) and \(f ( b ) > f ( c )\) or with \(f ( a ) > f ( b )\) and \(f ( b ) < f ( c )\) . In the first case, let m be the point at which \(f ( x )\) is maximized in \([ a , c ] ;\) such a point exists since f is continuous. Moreover we must have \(m \neq a , c\) by the hypothesis. But now the image of \(( a , c )\) under f contains \(f ( m )\) , but does not contain a neighborhood of \(f ( m )\) , so \(f\) cannot map open sets to open sets.
The second case is completely analogous.