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# 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 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 contains a contradiction. Thus the antithesis is wrong, and $f(\xi)=0$.

This means 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.

## Mathematics Subject Classification

26A15*no label found*

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