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Math Help - Continuous and Unbounded Function

  1. #1
    MHF Contributor harish21's Avatar
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    Continuous and Unbounded Function

    If a set S contains an unbounded sequence , show that the function f:S \rightarrow R defined by f(x)=x for all x in S, is continuous, but unbounded. and If a set S contains a sequence that converges to a point x_0 in S, show that the function f:S \rightarrow R defined by f(x) = \frac{1}{x-x_0} for all x in S, is continuous and unbounded

    All I could figure out was that
    for the first case,  S \equiv [0, \infty) contains an unbounded sequence. We can set, for every index n, {a_n} = n and thus every subsequence of this sequence is unbounded and fails to converge.

    How can it be proven for the defined function?

    Also in the second part, S \equiv (0, 2] has a sequence \frac{1}{n} that converges to 0 that is not in S.

    Again I am not being able to do this for the given function.

    Any suggestions??
    Last edited by harish21; March 25th 2010 at 08:15 AM.
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  2. #2
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    For the first part, we know there is some unbounded sequence \{s_n\} \subset S, so the sequence f(s_n) = s_n is obviously unbounded, meaning f is unbounded.

    To show that it's continuous, use the sequence definition of continuity; a function is continuous if s_n \rightarrow s implies f(s_n) \rightarrow f(s). That should be obvious since f is the identity function.

    The second part is similar.
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