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Math Help - Proof: Extreme Value Theorem

  1. #1
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    Proof: Extreme Value Theorem

    If f(x) is continuous in [a,b] there exists a point c in the closed interval [a,b] such that for all x in [a,b], f(x) cannot be greater than f(c). Also, there exists a point d in the interval [a,b] such that for all x in [a,b], f(x) cannot be less than f(d).

    One of the more complicated theorems I have come across, the Extreme Value Theorem (single-variable, real functions only) requires a proof that I don't see in my textbook.

    I've searched for it everywhere, and the only proof I have found is on wikipedia but the proof is to complicated for me to understand (I'm in 11th grade but I have been able to understand the other theorems' proofs no idea why this one should be such a stumper).

    Could someone sort of breakdown the reasoning of the proof in simpler terms in case that's possible?
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  2. #2
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    Quote Originally Posted by Skyrim View Post
    If f(x) is continuous in [a,b] there exists a point c in the closed interval [a,b] such that for all x in [a,b], f(x) cannot be greater than f(c). Also, there exists a point d in the interval [a,b] such that for all x in [a,b], f(x) cannot be less than f(d).
    Could someone sort of breakdown the reasoning of the proof in simpler terms in case that's possible?
    There is no simple proof.
    Proofs depend upon compactness, closed intervals are compact.
    It can be done using uniform continuity, continuous functions on compact set are uniformly continuous. We can show then that continuous functions on compact sets are bounded. Then if bounded they have a LUB and GLB each of which are limit points of the range of the function. Thus, there is a highpoint and a lowpoint in the range.

    Now that is no proof! But it tells you what is going on.
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  3. #3
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    Quote Originally Posted by Plato View Post
    There is no simple proof.
    Proofs depend upon compactness, closed intervals are compact.
    It can be done using uniform continuity, continuous functions on compact set are uniformly continuous. We can show then that continuous functions on compact sets are bounded. Then if bounded they have a LUB and GLB each of which are limit points of the range of the function. Thus, there is a highpoint and a lowpoint in the range.

    Now that is no proof! But it tells you what is going on.
    Actually I was wondering whether you could somehow prove it using the Intermediate Value Theorem. The two theorems look awfully similar, there must be some sort of connection that eludes my mind... or maybe not. Thanks for the help anyways. I was wondering if you could give me a link to a full proof of the theorem. I will try to work through it this summer.
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  4. #4
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    It's not a consequence of the IVT; the IVT holds on open intervals, for instance, where the EVT certainly fails. As Plato said, fundamentally it's about compactness, and that continuous images of compact sets are themselves compact, and that compact subsets of metric spaces are bounded.
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