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Math Help - proof dealing with supremum and bounds

  1. #1
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    proof dealing with supremum and bounds

    I'm having trouble understanding how to prove this problem. I need some help.

    Let x Real number and Let S be a non empty subset of real numbers that is bounded above. We define a new set x + S, by x+S={x+s: s S}

    a) Prove that x +S is bounded above
    b) Prove that x + sup(S) is an upper bound of x + S. Conclude that the sup(x+ S) ≤ x + sup(S)
    C) Prove that x + sup(S) = sup(S) = sup(x+S)

    Thanks for any help provided.
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  2. #2
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    Quote Originally Posted by tuyt6444 View Post
    I'm having trouble understanding how to prove this problem. I need some help.

    Let x Real number and Let S be a non empty subset of real numbers that is bounded above. We define a new set x + S, by x+S={x+s: s S}

    a) Prove that x +S is bounded above
    Suppose p is an upper bound for S. Then y\le p for all y in S. What can you say about y+ S?

    b) Prove that x + sup(S) is an upper bound of x + S. Conclude that the sup(x+ S) ≤ x + sup(S)
    C) Prove that x + sup(S) = sup(S) = sup(x+S)
    This last statement is clearly untrue. Do you really have that "sup(S)" in the middle?

    Thanks for any help provided.
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  3. #3
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    Sorry I made a mistake on that last one the correct problem is

    (c) prove that x+sup(S) = sup(x+S)
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