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Math Help - if f:A->B with A closed & bouned, then f(A) is closed & bounded.

  1. #1
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    if f:A->B with A closed & bouned, then f(A) is closed & bounded.

    I have written a valid proof that f(A) is bounded. I don't know how to show that f(B) is also closed.

    I am wondering if I can use the Completeness Theorem to argue that there is a supremum and infimum. Then I would need to show that those two numbers reside in A for A to be closed and not open. I don't know how to do that, but am I at least on the right track?

    Nevermind, I found out that I am on the right track, and that I just need to use the Maximum-Minimum Theorem.
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  2. #2
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    Quote Originally Posted by anon2194 View Post
    I have written a valid proof that f(A) is bounded. I don't know how to show that f(B) is also closed.

    I am wondering if I can use the Completeness Theorem to argue that there is a supremum and infimum. Then I would need to show that those two numbers reside in A for A to be closed and not open. I don't know how to do that, but am I at least on the right track?

    Nevermind, I found out that I am on the right track, and that I just need to use the Maximum-Minimum Theorem.

    And, of course, you must assume f is continuous...

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by anon2194 View Post
    I have written a valid proof that f(A) is bounded. I don't know how to show that f(B) is also closed.

    I am wondering if I can use the Completeness Theorem to argue that there is a supremum and infimum. Then I would need to show that those two numbers reside in A for A to be closed and not open. I don't know how to do that, but am I at least on the right track?

    Nevermind, I found out that I am on the right track, and that I just need to use the Maximum-Minimum Theorem.
    I assume since you are discussing the completeness theorem that you are referring to A,B\subset\mathbb{R}. How exactly did you prove boundedness without proving closure? Did you use the fact that A is compact?
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