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.