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

    All I could figure out was that
    for the first case, $\displaystyle S \equiv [0, \infty)$ contains an unbounded sequence. We can set, for every index n, $\displaystyle {a_n} = n$ 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, $\displaystyle S \equiv (0, 2]$ has a sequence $\displaystyle \frac{1}{n}$ 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; Mar 25th 2010 at 07:15 AM.
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  2. #2
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    For the first part, we know there is some unbounded sequence $\displaystyle \{s_n\} \subset S$, so the sequence $\displaystyle f(s_n) = s_n$ is obviously unbounded, meaning f is unbounded.

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

    The second part is similar.
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