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Math Help - Proving a supremum exists.

  1. #1
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    Proving a supremum exists.

    The question is:

    Let A = {p/q : p,q are positive integers and p < 25q }
    Prove, using the definition of the supremum, that supA = 25.

    I have proved that \alpha = 25, which is an upper bound. What remains to prove is that it is the least upper bound and this is done on the answer sheet by reductio ad absurdum.

    The answer sheet reads "Suppose that \gamma is an upper bound of A and \gamma < \alpha. By the Archimedian property of the rational numbers, there is an x= p/q \in Q such that

    \gamma < x < 25 = \alpha

    Since we can assume both p and q are positive integers we see that x < 25 implies that p < 25q. Therefore x \in A and x > \gamma. Therefore \gamma is not an upper bound. Therefore, if \gamma is an upper bound of A, then \gamma \ge \alpha. Thus \alpha = 25 is the supremum."

    I draw your attention to the inequality on a line on its own. (I couldn't seem to make the symbols bold.) Surely if \gamma is an upper bound to A, it is by definition greater than all x in A?



    A very confused mathematician.
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  2. #2
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    That proof is correct.
    Here is what it shows. \alpha is an upper bound and no number less than \alpha is an upper bound, therefore \alpha is the least upper bound, supremum.
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  3. #3
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    Ah, thank you for clarifying things. I think I was approaching it from the wrong angle. Is it saying that no x can exist that will fulfil the condition \gamma < x < 25 = \alpha whilst the condition that \gamma is an upper bound, so \gamma cannot exist?
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  4. #4
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    Quote Originally Posted by iamahorse View Post
    Ah, thank you for clarifying things. I think I was approaching it from the wrong angle. Is it saying that no x can exist that will fulfil the condition \gamma < x < 25 = \alpha whilst the condition that \gamma is an upper bound, so \gamma cannot exist?
    Yes, that is what I said.
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