0

Let $f(x):\mathbb R\to \mathbb R$ be a continuous function. How to prove that $f$ can have at most countably many strict local maxima?

I've been thinking about choosing a rational point for each local maxima in $(x_0 -\delta;x_0 + \delta)$ where $f(x_0) > f(x)$ for any $x \in (x_0 -\delta;x_0 + \delta)$. But how to do that so for different $x_1$ and $x_2$ stand different points?

William Elliot
  • 17,598
Mark Tiukov
  • 87
  • @postmortes, I don't get there last part: The set of strict local maxima is a countable union of such sets, for example taking $\delta=\frac{1}{n}$ as n ranges over the positive integers. – Mark Tiukov Dec 19 '19 at 11:44
  • why we can say so? – Mark Tiukov Dec 19 '19 at 11:44
  • since ${\mathbb N}$ is countable, $1/n$ for $n \in {\mathbb N}$ is also countable: the map $f: n \rightarrow 1/n$ is one-to-one so preserves the countability. In fact, any one-to-one transformation for your $\delta$ will preserve countability – postmortes Dec 19 '19 at 11:49

0 Answers0