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Math Help - Extreme Value Theorem

  1. #1
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    Extreme Value Theorem

    Here is a proof from my notes that I could use some help with.

    Let I and J be closed intervals. Suppose f:I x J\rightarrow R is continuous. Show there are (x_{min},y_{min}) and (x_{max},y_{max}) in I x J, such that f(x_{min},y_{min})\leq f(x,y)\leq f(x_{max},y_{max}), for all (x,y)\in I x J.

    The hint that we are given is to divide the rectangle I x Jinto four subrectangles. Then, the supremum of f over at least one of these subrectangles must be equal to the supremum of f over I x J...

    I see that they are just applying the Extreme Value Theorem to rectangles but I'm not sure how to word the proof.
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  2. #2
    Senior Member Tinyboss's Avatar
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    The short answer is that the product of compact sets is compact, and a continuous function with compact domain has such points.

    But I think you're probably building up to that theorem. I think the idea here is to continue the process: take the rectangle where the supremum of f occurs, and divide it again, and so on. You'll then be able to define a Cauchy sequence in IxJ, which converges since ... and the value of f at the limit point is ...
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  3. #3
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    Ok, so I just divide the rectangles into smaller and smaller rectangles until I have a rectangle that contains only one point, x_{max}. So, x_{max} is the limit point. But then I'm not really sure what else to say. I'm guessing that we have to verify that this point x_{max} is actually the supremum of the entire rectangle.
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  4. #4
    Senior Member Tinyboss's Avatar
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    Well, you don't ever reach a rectangle with only one point. But there is a well-defined limit to the process. Now what can you say, for f continuous, about lim f(x) as x approaches some point?
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  5. #5
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    Since f is continuous then the lim_{x\rightarrow x_{max}}f(x)=f(x_{max}})
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