Preservation of Intervals Theorem

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

Preservation of Intervals Theorem

We will now look at two important theorems regarding the preservation of intervals under a function. Provided that $f : I \to \mathbb{R}$ is a continuous function over the closed and bounded interval $I$, then $f(I)$ will also be a closed and bounded interval, and more generally, if $I$ is any interval, then $f(I)$ will also be an interval.

 Theorem 1: If $f : I \to \mathbb{R}$ is a continuous function on $I$ and if $I = [a, b]$ is a closed and bounded interval, then the range of $f$, $f(I) = \{ f(x) : x \in I \}$ is also a closed and bounded interval. • Proof: Let $f : I \to \mathbb{R}$ be a continuous function on the closed and bounded interval $I$ into the set of real numbers. Since $I$ is a closed and bounded interval, then there exists $M, m \in \mathbb{N}$ such that $M = \sup f(I)$ and let $m = \inf f(I)$. By The Maximum-Minimum Theorem, $m$ is the absolute minimum of $f$ and $M$ is the absolute maximum of $f$. Therefore we have that $m ≤ f(x) ≤ M$ for all $x \in I$ and so $f(I) \subseteq [m, M]$.
• We now want to show that $[m, M] \subseteq f(I)$. Let $k \in [m, M]$. Then by the corollary to Bolzano's Intermediate Value Theorem there exists a $c \in I$ such that $f(c) = k$. Therefore $f(c) = k \in f(I)$ and so $[m, M] \subseteq f(I)$.
• Therefore $f(I) = [m, M]$, and so $f(I)$ is a closed and bounded interval. $\blacksquare$

We will now look at the Preservation of Intervals Theorem which is a generalization of theorem 1.

 Theorem 2 (Preservation of Intervals): If $f : I \to \mathbb{R}$ is a continuous function on $I$, and if $I$ is an interval, then the range of $f$, $f(I) = \{ f(x) : x \in I \}$ is also an interval. • Proof: Let $f : I \to \mathbb{R}$ be a continuos function from the interval $I$ into the set of real numbers, and let $\alpha, \beta \in f(I)$ be such that $\alpha < \beta$. Then there exists $a, b \in I$ such that $f(a) = \alpha$ and $f(b) = \beta$. From Bolzano's Intermediate Value Theorem, if $k \in (\alpha, \beta)$, then there exists a $c \in I$ where $k = f(c) \in f(I)$. Therefore since $\alpha, \beta \in f(I)$ and $[\alpha, \beta] \subseteq f(I)$, by The Characterization Theorem for Intervals theorem, $f(I)$ is an interval. $\blacksquare$