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    Upper bounds of a non-empty set

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    Let S be a non-empty set of real numbers, and prove that the following statements are equivalent:

    (1) If v is any upper bound of S, then u <= v (read as "u is less than or equal to v").
    (2) If z < u, then z is not an upper bound of S.
    (3) If z < u, then there exists s_z (read as "s sub z") in S such that z < s_z.
    (4) If epsilon > 0, then there exists s_epsilon (read as "s sub epsilon") in S such that u - epsilon < s_epsilon.

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    Solution Preview

    To prove that statements (1) through (4) are equivalent, we will prove the following:

    (1) -> (2)

    (2) -> (3)

    (3) -> (4)

    (4) -> (1)

    --------------------------------------------------------------

    Proof that (1) -> (2):

    Assume that (1) holds, and prove that (2) holds.

    Suppose (by way of contradiction) that (2) does not hold. Then there exists z < u such that z is an upper bound of S. Setting v to z in (1), we obtain u <= z, since z is an upper bound of S, but then we have u <= z < u, which is absurd, since the statements "u <= z" and "u > z" cannot both be true. Thus our supposition that (2) does not hold is false, so (2) ...

    Solution Summary

    A detailed proof of the equivalence of the four given statements is provided. Specifically, the proof consists of showing the following: (a) statement (1) implies statement (2); (b) statement (2) implies statement (3); (c) statement (3) implies statement (4); (d) statement (4) implies statement (1).

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