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  $f(a)\ne 0$. 

The set  $A=\{x\in [a,b]\mathrm{⋮}f(x)=0\}$  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  $f(\xi )\ne 0$.  Then, by the continuity of $f$, there is a positive number $\delta$ such that

Chose a number $x_1$ between $\xi$ and $\xi +\delta$; then  $f(x_1)\ne 0$,  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 $x_1$ ().  Now  ,  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.