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Proof about the decomposition of the reals into two sets.

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Let S and T be nonempty sets of real numbers such that every real number is in S or T and if s [itex]\in[/itex] S and t [itex]\in[/itex] T, then s < t. Prove that there is a unique real number β such that every real number less than β is in S and every real number greater than β is in T.

    3. The attempt at a solution
    I tried a proof by contradiction, but I started with the assumption that the preposition was true (not sure if that is OK). I haven't had a formal introduction into proof-writing. I'm 3 weeks into my calculus course and that was one of the exercises given in the TB.

    The preposition states that S has a sup β, and T has an inf β, where β is a unique real number.

    Suppose that β does not exist. Thus S does not have a supremum, and is not bounded above. T does not have an infimum, and is not bounded below. Thus both S and T are the set of real numbers.

    Therefore, there exists an so and a to such that So > To.

    I would really appreciate comments because like I said I'm completely new to this stuff. :(
     
  2. jcsd
  3. Mar 5, 2012 #2

    HallsofIvy

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    Do you have theorems about "greatest lower bound" or "least upper bounds" of sets of numbers? Such as "if a set of real numbers has an upper bound then it has a least upper bound (supremum)" or "if a set of real numbers has a lower bound then it has a greatest lower bound (infimum)". Those are what you need.
     
  4. Mar 5, 2012 #3
    We know that T is bounded below and S is bounded above - how?

    Then we can get onto comparing least/greatest bounds as applicable...
     
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