# least and greatest zero

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).  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 LeastAndGreatestZero 2013-03-22 16:33:22 2013-03-22 16:33:22 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 26A15 zeroes of continuous function ZeroesOfAnalyticFunctionsAreIsolated