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Proof of Supremum of a Set

  1. May 26, 2014 #1
    The problem statement, all variables and given/known data

    8. Let ##A## be a non-empty subset of ##R## which is bounded above. Define
    ##B = \{x ∈ R : x − 1 ∈ A\}##, ##C = \{x ∈ R : (x + 1)/2 ∈ A\}.##
    Prove that sup B = 1 + sup A, sup C = 2 sup A − 1.



    The attempt at a solution

    Note that ##sup A## exists. Let ##x ∈ B##; then ##x − 1 ∈ A##, and ##x − 1 ≤ sup A##. We have that ##x ≤ sup A + 1##, from which we deduce that ##B## is bounded above and ##sup B ≤ sup A + 1##. Now suppose that ##m## is an upper bound for ##B##. For ##x ∈ A##, ##x + 1 ∈ B## and ##x + 1 ≤ m##. It follows that ##x ≤ m − 1##, from which we deduce that ##m − 1## is an upper bound for ##A## and ##sup A ≤ m − 1##. Now ##sup A + 1 ≤ m##. Combining the above, ##sup B = sup A + 1##.

    I understand this solution up until we reach ##sup B = sup A + 1##. I understand that they both must be less than or equal to ##m##, as ##m## is an upper bound for ##B## and an upper bound for ##A + 1##. However I don't understand how we know that they are exactly equal to each other. Could anyone help me out please?
     
    Last edited by a moderator: May 26, 2014
  2. jcsd
  3. May 26, 2014 #2

    verty

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    ##m## is arbitrary, it can be any upper bound, even the least upper bound.
     
    Last edited: May 26, 2014
  4. May 26, 2014 #3

    Fredrik

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    LaTeX tip: \{x\in\mathbb R:x-1\in A\}
     
  5. May 26, 2014 #4

    micromass

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    I fixed the LaTeX in the OP
     
  6. May 26, 2014 #5

    HallsofIvy

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    First, since A is bounded above, so is B and so B has a supremum. Let [itex]\alpha[/itex] be the supremum of A.

    1) Suppose [itex]x> \alpha+ 1[/itex]. Then [itex]x- 1>\alpha[/itex]. Since [itex]\alpha[/itex] is an upper bound on A x- 1 is not in A. Therefore, x= (x- 1)+ 1 is not in B. That is [itex]\alpha+ 1[/itex] is an upper bound on B.

    2) Suppose there exist y, an upper bound on B, with [itex]y< \alpha+ 1[/itex]. Then if [itex]y< x< \alpha+ 1[/itex] x is not in B. But then [itex]x- 1< \alpha[/itex] is not in A which contradicts the fact that [itex]\alpha[/itex] is the least upper bound on A.
     
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