least and greatest zero

Theorem. If a real function $f$ is continuous on the interval $[a,\,b]$ and has zeroes on this interval, then $f$ has a least zero and a greatest zero.

Proof. If $f(a)=0$ then the assertion concerning the least zero is true. Let’s assume therefore, that $f(a)\neq 0$ .

The set $A=\{x\in[a,\,b]\vdots\,\,f(x)=0\}$ is bounded from below since all numbers of $A$ are greater than $a$ . Let the infimum (http://planetmath.org/InfimumAndSupremumForRealNumbers) of $A$ be $\xi$ . Let us make the antithesis, that $f(\xi)\neq 0$ . Then, by the continuity of $f$ , there is a positive number $\delta$ such that

f(x)\neq 0\quad\mathrm{always\,when}\,\,|x-\xi|<\delta.

Chose a number $x_{1}$ between $\xi$ and $\xi\!+\!\delta$ ; then $f(x_{1})\neq 0$ , but this number $x_{1}$ is not a lower bound of $A$ . Therefore there exists a member $a_{1}$ of $A$ which is less than $x_{1}$ ( $\xi<a_{1}<x_{1}$ ). Now $|a_{1}-\xi|<|x_{1}-\xi|<\delta$ , whence this member of $A$ ought to satisfy that $f(a_{1})=0$ . This a contradiction. Thus the antithesis is wrong, and $f(\xi)=0$ .

This that $\xi\in A$ and $\xi$ is the least number of $A$ .

Analogically one shows that the supremum of $A$ is the greatest zero of $f$ on the interval.

Title	least and greatest zero
Canonical name	LeastAndGreatestZero
Date of creation	2013-03-22 16:33:22
Last modified on	2013-03-22 16:33:22
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Theorem
Classification	msc 26A15
Synonym	zeroes of continuous function
Related topic	ZeroesOfAnalyticFunctionsAreIsolated