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Thread: 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= $\displaystyle \infty$
    how can we show that?
    $\displaystyle S= \{f(x):\int_0^T f(x) dx = 1\}$
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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= $\displaystyle \infty$
    how can we show that?
    $\displaystyle S= \{f(x):\int_0^T f(x) dx = 1\}$
    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 $\displaystyle S$ is closed (for $\displaystyle T\in \field{R}$) you show that the limit $\displaystyle f$ of any convergent sequence of functions $\displaystyle f_n\in S$ also lies in $\displaystyle S$.

    2. To show that $\displaystyle S$ is not closed (for $\displaystyle T=\infty$) you show that there exists a sequence of functions $\displaystyle f_n$ in $\displaystyle S$, the limit $\displaystyle f$ of which does not lie in $\displaystyle S$. (For example: it is easy to find a sequence of functions $\displaystyle f_n\in S$ with limit $\displaystyle f \equiv 0,\notin S$.)
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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 $\displaystyle S = \{x : \int_0^T f = 1\}$, specifically, is closed? I assume so. (Give enough detail with your problem!)

    I suppose the integral is the usual Riemann integral and $\displaystyle f$ is Riemann-integrable on every compact subset of $\displaystyle \mathbb{R}$. Then $\displaystyle F(t)=\int_0^T f$ is continuous, and for any continuous function $\displaystyle F: \mathbb{R}\to \mathbb{R}$, and any $\displaystyle \alpha \in \mathbb{R}$, the set $\displaystyle \{x : F(x)=\alpha\}$ 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 $\displaystyle L^2$ space of square-integrable functions.
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