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Proving on the completeness theorem of real number

  1. Aug 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Let A be a non empty subset of R that is bounded above

    Set D:={2a|a (belongs to) A}
    Is it necessarily true that the sup D = 2 sup A? Either prove or provide a counterexample.

    2. Relevant equations
    The completeness axiom

    3. The attempt at a solution
    I am seriously clueless on how to approach... but I still tried something
    Let sup D = y and sup A = x
    d=2a; d<=y; a<=x

    or can I say choose a as the largest value in A, so 2a=d is the largest value in D. Since both are the upper bound for each set and for all upper bound and real number, they are the smallest. so, d is sup D and a is sup A.
    Therefore, d=2a => sup D = 2 sup A

    But these method seems a bit weird...
  2. jcsd
  3. Aug 6, 2011 #2


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    Science Advisor

    No, you can't say "choose a as the largest value in A". The whole point of "sup" is that the set may not have a largest value. And you can't say "Let sup D = y and sup A = x d=2a; d<=y; a<=x" because you didn't say what "a" was.

    "sup" has two properties- it is an upper bound (so there are no member of the set larger than the sup) and there is no smaller upper bound (so there are members of the set within any "[itex]epsilon[/itex]" distance of the sup.

    Let S be the sup of set A. We can prove that 2S is an upper bound for 2A by contradiction:
    suppose there exist b in 2A such that b> 2S. But b= 2a for some a in A. That says 2a> 2S so that a> S which contradicts the fact that S is an upper bound for A.

    Similarly, you can prove by contradiction that there is no upper bound less than 2S.
  4. Aug 6, 2011 #3
    But how do we relate this to the prove of sup D = 2 sup A?

    Does 2S means D?

    Sorry, I'm a bit confused now
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