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Math Help - open and closed set

  1. #1
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    open and closed set

    how does one go about showing that a set is open/closed.

    e.g the set below is said tobe closed for finite values of T. but not closed for T= \infty
    how can we show that?
    S= \{f(x):\int_0^T f(x) dx = 1\}
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by mtmath View Post
    how does one go about showing that a set is open/closed.

    e.g the set below is said tobe closed for finite values of T. but not closed for T= \infty
    how can we show that?
    S= \{f(x):\int_0^T f(x) dx = 1\}
    First, you need to be perfectly clear about the topological space in which this whole problem statement is to be situated.

    If it is a metric space you can do the following:

    1. To show that S is closed (for T\in \field{R}) you show that the limit f of any convergent sequence of functions f_n\in S also lies in S.

    2. To show that S is not closed (for T=\infty) you show that there exists a sequence of functions f_n in S, the limit f of which does not lie in S. (For example: it is easy to find a sequence of functions f_n\in S with limit f \equiv 0,\notin S.)
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    I'm not sure what the problem is asking for specifically. In general, to show a set is closed, you want to show its complement is open. Are you asking whether and why the set S = \{x : \int_0^T f = 1\}, specifically, is closed? I assume so. (Give enough detail with your problem!)

    I suppose the integral is the usual Riemann integral and f is Riemann-integrable on every compact subset of \mathbb{R}. Then F(t)=\int_0^T f is continuous, and for any continuous function F: \mathbb{R}\to \mathbb{R}, and any \alpha \in \mathbb{R}, the set \{x : F(x)=\alpha\} is closed. (Prove this! Just apply the definition of continuity.)
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  4. #4
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    hi,
    the set S is defined in the L^2 space of square-integrable functions.
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