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Explanation of the Extreme Value Theorem proof

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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.

Re: Explanation of the Extreme Value Theorem proof

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?

Re: Explanation of the Extreme Value Theorem proof

Nono, f(x) cannot be just less than equal M. Why? If $\displaystyle M = Sup(f(x))_{x \in I } $. Then since X is a closed and bounded domain and f is continuous, Then it means that there exists a sequence $\displaystyle x_n \in X $ such that $\displaystyle x_n \to a $ implies $\displaystyle f(x_n) \to M $. Since X is closed and bounded, X contains a, so this must mean that $\displaystyle f(x_n) \to M = f(a) $. So there is a point a where it reaches the sup so it cannot be less than.

Also $\displaystyle M = Sup(f(x))_{x \in I } $ . So What he shows is that $\displaystyle f(x) \leq M - \frac{1}{C} $ and $\displaystyle sup(f(x))_{x \in I } = M \leq M - \frac{1}{C}$ which means M < M

Re: Explanation of the Extreme Value Theorem proof

Quote:

Originally Posted by

**coolge** 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?

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))