You have f(x) <= M - (1/C) for all x in I. Sup f(x) = M <= M - 1/C <M since C>0 Hence M <M.
Am I missing something in your question?
In the proof to this theorem, how does taking the supremum of (8), yield the contradicting inequality? We know from the premises that f(x) < M, for all x in I. However, the statement f(x) <= M - (1/C) for all x in I does not in any way lead to the contradiction stated in the proof. Here is my reasoning:
f(x) < M for all x in I. f(x) <= M - (1/C). So, taking the second statement into consideration, it is less than or equal to, but the first condition holds in either less than or equal to.
Nono, f(x) cannot be just less than equal M. Why? If . Then since X is a closed and bounded domain and f is continuous, Then it means that there exists a sequence such that implies . Since X is closed and bounded, X contains a, so this must mean that . So there is a point a where it reaches the sup so it cannot be less than.
Also . So What he shows is that and which means M < M
No, but I don't understand how you can set up an inequality such as "Sup f(x) = M <= M - 1/C <M". What I understand is that if you start with the premise that f(x) < M = sup(f(x)) for all x in I, then you end up with a contradiction because, if this is the case, then M - (1/C) is also the sup(f(x))