Results 1 to 4 of 4

Math Help - Explanation of the Extreme Value Theorem proof

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    20

    Explanation of the Extreme Value Theorem proof

    Explanation of the Extreme Value Theorem proof-extreme-value-theorem.gif
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2012
    From
    NY
    Posts
    62
    Thanks
    8

    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    388
    Thanks
    80

    Re: Explanation of the Extreme Value Theorem proof

    Nono, f(x) cannot be just less than equal M. Why? If  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  x_n \in X such that  x_n \to a implies  f(x_n) \to M . Since X is closed and bounded, X contains a, so this must mean that  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  M = Sup(f(x))_{x \in I } . So What he shows is that  f(x) \leq M - \frac{1}{C} and  sup(f(x))_{x \in I } = M \leq M - \frac{1}{C} which means M < M
    Last edited by jakncoke; December 2nd 2012 at 05:39 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2011
    Posts
    20

    Re: Explanation of the Extreme Value Theorem proof

    Quote Originally Posted by coolge View Post
    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))
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof: Extreme Value Theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 24th 2011, 06:29 PM
  2. Extreme Value Theorem
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 8th 2010, 05:34 AM
  3. Extreme Value Theorem
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: March 27th 2010, 06:04 PM
  4. Extreme Value Theorem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 12th 2009, 10:28 PM
  5. Applying the Extreme Value Theorem
    Posted in the Calculus Forum
    Replies: 9
    Last Post: February 9th 2008, 03:14 PM

Search Tags


/mathhelpforum @mathhelpforum