We denote by $\widehat{S}$ the set of $f \in S_*$ satisfying $\lim_{x \rightarrow -\infty} f(x) = -\infty$ and $\lim_{x \rightarrow +\infty} f(x) = +\infty$. We denote by $\operatorname{Mi}(f)$ the set of minima of $f$ and by $\operatorname{Ma}(f)$ the set of maxima of $f$, so $E(f) = \operatorname{Mi}(f) \cup \operatorname{Ma}(f)$. 1. Let $f \in \widehat{S}$. a. Verify that $\operatorname{Card} \operatorname{Mi}(f) = \operatorname{Card} \operatorname{Ma}(f)$ and that for $y \in \mathbb{R}$, $f^{-1}([-\infty, y])$ is the union of non-empty open intervals that are pairwise disjoint. We denote by $\mathscr{I}(y)$ their set. b. Show that for every element $M$ of $\operatorname{Ma}(f)$, there exists a unique element $m$ of $\operatorname{Mi}(f)$ such that $f(m) < f(M)$ and $m > M$.
We denote by $\widehat{S}$ the set of $f \in S_*$ satisfying $\lim_{x \rightarrow -\infty} f(x) = -\infty$ and $\lim_{x \rightarrow +\infty} f(x) = +\infty$. We denote by $\operatorname{Mi}(f)$ the set of minima of $f$ and by $\operatorname{Ma}(f)$ the set of maxima of $f$, so $E(f) = \operatorname{Mi}(f) \cup \operatorname{Ma}(f)$.
1. Let $f \in \widehat{S}$.\\
a. Verify that $\operatorname{Card} \operatorname{Mi}(f) = \operatorname{Card} \operatorname{Ma}(f)$ and that for $y \in \mathbb{R}$, $f^{-1}([-\infty, y])$ is the union of non-empty open intervals that are pairwise disjoint. We denote by $\mathscr{I}(y)$ their set.\\
b. Show that for every element $M$ of $\operatorname{Ma}(f)$, there exists a unique element $m$ of $\operatorname{Mi}(f)$ such that $f(m) < f(M)$ and $m > M$.