determining inf S and sup S?

Let S = { (1/n) - (1/m) } such that n, m are in the set of natural numbers. Determine inf S and sup S. Justify your answers

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  • kb
    Lv 7
    9 years ago
    Favorite Answer

    inf S = -1, and sup S = 1.

    For inf S, note that 1/n - 1/m < 0 - 1/1 = -1 for all m, n in N.

    Hence, -1 is a lower bound for S.

    To show that S is the greatest lower bound of S, let ε > 0 be given and consider -1 + ε.

    We show that -1 + ε is not a lower bound for S.

    -By Archimedian Property, there exists n in N such that nε > 1 <==> ε > 1/n.

    Hence, -1 + ε > -1 + 1/n = 1/n - 1/1. (So, we've found an element in S smaller than -1 + ε.)

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    The proof for sup S is similar.

    Note that 1/n - 1/m < 1/1 - 0 = 1 for all m, n in N.

    Hence, 1 is an upper bound for S.

    To show that S is the least upper bound of S, let ε > 0 be given and consider 1 - ε.

    We show that 1 - ε is not a lower bound for S.

    -By Archimedian Property, there exists m in N such that mε > 1 <==> -ε < -1/m.

    Hence, 1 - ε < 1 - 1/m. (So, we've found an element in S greater than 1 - ε.)

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    I hope this helps!

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