Deducing the "Maximum Theorem" from the 'Boundedness Theorem' by an indirect argument.

Question

I'm solving a problem says that;

Deduce the "Maximum Theorem" from the 'Boundedness Theorem' by an indirect argument. Namely, assume $f(x)$ has no maximum on $I$ ; let $M = \sup _{I} f(x)$, apply the Boundedness Theorem to $\frac {1} {M-f(x)}$.

I know both boundedness theorem and minimum-maximum theorem. But I don't know what the hack of this problem wants me to do.

First, $M$ is also upper-bound of $f(x)$, hence $M \ge f(x)$, so $M - f(x) \ge 0 $, leading

$0 < \frac {1} {M-f(x)} < \infty.$

So now what? I can't catch any clue between compactness or continuity and the hell of that.

It will be very so much pleasure if you would give me some of help. Thank you for all this.

Answer 1

Assuming that $f$ has no maximum you want to prove that it is not bounded, which means $M=\infty$. You prove this by contradiction. Suppose $M<\infty$. Consider $\frac 1 {M-f(x)}$. If $C$ is a bound for this function then $\frac 1 {M-f(x)} \leq C$ or $ f(x)<M-\frac1 C $ for all $x$. But then $M-\frac 1 C$ is an upper bound for $F$ which contradicts the fact that $M$ is the least upper bound.