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. $\blacksquare$