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Math Help - Intro to Real Analysis Proof

  1. #1
    Member ilikedmath's Avatar
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    Question Intro to Real Analysis Proof

    I am really having a hard time in this intro to real analysis class. I feel as if I'm the only one in class who isn't getting it. I have an extremely hard time thinking abstractly and constructing my own proofs. I know I need a lot of practice. Here is the problem we have to prove:


    Claim: Let A be a nonempty subset of R (all real numbers -- how do I type the symbol for real numbers?). If α = sup A is finite, show that for each ε > 0, there is an a in A such that α – ε < a ≤ α.

    My attempt of a proof: Assume α = sup A is finite. Then A is bounded above because it is not empty and its supremum is finite (by the definition that if E is a nonempty subset of R (all reals), we set sup E = ∞ if E is not bounded above). [my question is where does the “ε” come from?] By definition of supremum, there is an element in R such that < α and is not an upper bound. In this case let ε be the where ε > 0. Knowing α is the supremum, ε < α, so there is an element a in A such that ε < a ≤ α or α – ε < a ≤ α.

    *I also need to prove the converse of this statement which is:
    "Let A be a nonempty subset of R (all real numbers) that is bounded above by α. Prove that if for every ε > 0 there is an a in A such that α – ε < a ≤ α, then α = sup A."

    When proving the converse, isn't it just basically working backwards?
    So I would write: Assume that for every ε > 0 there is an a in A such that α – ε < a ≤ α.
    A is nonempty and bounded above by α (given). Then α = sup A is finite by the definition of supremum.

    I feel really confused and lost here. I'm really afraid of this class. I need to pass it because it is only offered every 2 years.

    Any help, suggestions, and guidance is greatly appreciated.
    Thank you.
    Last edited by ilikedmath; September 28th 2008 at 03:14 PM. Reason: Forgot to include converse
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  2. #2
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    First. What does it mean to say that x is an upper bound for set A?
    Second. What does it mean to say y is not an upper bound of set A?
    Third. What does it mean to say z is the least upper bound (the sup) of set A?
    Finally. If z=sup(A) is it possible that z - \varepsilon is an upper bound if \varepsilon > 0?
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  3. #3
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    Quote Originally Posted by ilikedmath View Post
    Claim: Let A be a nonempty subset of R (all real numbers -- how do I type the symbol for real numbers?). If α = sup A is finite, show that for each ε > 0, there is an a in A such that α ε < a ≤ α.
    If a is the supremum then a-\epsilon cannot be the maximum since it is smaller than a and a is the supremum. Therefore there is x\in A so that a-\epsilon < x \leq a.
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  4. #4
    Member ilikedmath's Avatar
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    Question My work so far

    Here is what I have so far:

    Proof for 1st statement:

    Let A be a nonempty subset of R. Assume alpha = sup A is finite. For any a in A, a < alpha by the definition of supremum. Since alpha is the supremum, alpha - epsilon cannot be an upper bound of A given any epsilon > 0. Since alpha - epsilon is not an upper bound, there exists an a in A such that
    a > alpha - epsilon. Thus (alpha - epsilon) < a ≤ alpha. QED.

    To prove the converse of the 1st statement should've been simply working backwards, right? I got stuck though:

    Proof of converse:
    Let A be a nonempty subset bounded above by alpha. Assume for every epsilon > 0 there is an a in A such that (alpha - epsilon) < a ≤ alpha. Clearly (alpha - epsilon) < alpha, and so (alpha - epsilon) is not an upper bound.

    Now where can I go from there? (That is, if I am even on the right track)
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