extreme value theorem

Extreme Value Theorem. Let $a$ and $b$ be real numbers with , and let $f$ be a continuous, real valued function on $\mathrm{[}a\mathrm{,}b\mathrm{]}$. Then there exists $c\mathrm{,}d\mathrm{\in }\mathrm{[}a\mathrm{,}b\mathrm{]}$ such that $f\mathit{}\mathrm{(}c\mathrm{)}\mathrm{\le }f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\le }f\mathit{}\mathrm{(}d\mathrm{)}$ for all $x\mathrm{\in }\mathrm{[}a\mathrm{,}b\mathrm{]}$.

Proof. We show only the existence of $d$. By the boundedness theorem $f([a,b])$ is bounded above; let $l$ be the least upper bound of $f([a,b])$. Suppose, for a contradiction, that there is no $d\in [a,b]$ such that $f(d)=l$. Then the function

$$g(x)=\frac{1}{l-f(x)}$$

is well defined and continuous on $[a,b]$. Since $l$ is the least upper bound of $f([a,b])$, for any positive real number $M$ we can find $\alpha \in [a,b]$ such that $f(\alpha )>l-\frac{1}{M}$, then

So $g$ is unbounded on $[a,b]$. But by the boundedness theorem $g$ is bounded on $[a,b]$. This contradiction finishes the proof.