Extreme value theorem: help with contradiction

Question

I have a problem understanding the last part of the usual proof of the extreme value theorem (found for example here: Extreme Value Theorem proof help)

It is this part that I have trouble understanding:

Since $g(x)=\dfrac1{M−f(x)}\leq K$ is equivalent to $f(x)\leq M−\dfrac1K$, we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on $[a,b]$. Hence, there must be a balue $c\in[a,b]$ such that $f(c)=M$.

Why is it that we are sure that there exist a c on the interval where the maximum is attained? I mean, don't we just know that the new sup is $M-\dfrac1K$, but that it necessarily will not attain a max on the interval? Or do we have to assume that the function $g$ that we create attains a maximum for $f$ to attain a maximum?

Answer 1

Because we don't have a new $\sup f$. By definition, $M=\sup f$ and that proof proves that if there was no such $c$, then we would have $\bigl(\forall x\in[a,b]\bigr):f(x)\leqslant M-\frac1K$, which is impossible, because it would follow from that that $\sup f\leqslant M-\frac1K<M=\sup f.$$$

Answer 2

We have $M= \sup f([a,b])$. If we assume that there is no $c \in [a,b]$ with $f(c)=M$, then we have $f(x) <M$ for all $x \in [a,b]$ . In the above proof this leads to $f(x) \le M- \frac{1}{K}$ for all $x \in [a,b]$. Hence

$M = \sup f([a,b]) \le M- \frac{1}{K}$, a contradiction.