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Math Help - Annihilator of a set M in inner product space X

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    Annihilator of a set M in inner product space X

    Show that the annihilator (M perpendicular) of a non-empty set M in an inner product space X, is a closed subspace of X.
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    Quote Originally Posted by frater_cp View Post
    Show that the annihilator (M perpendicular) of a non-empty set M in an inner product space X, is a closed subspace of X.

    Let x,y \in M \perp

    Note to be in  M \perp <x,m> =<y,m> = 0, \forall m \in M

    Now lets show that it is closed. let \alpha, \beta \in \mathbb{R}

    <\alpha x+\beta y,m>=<\alpha x,m> +<\beta y,m>=\alpha <x,m>+\beta <y,m>=\alpha \cdot 0 +\beta \cdot 0 =0
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    Super Member Rebesques's Avatar
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    I think the question demands "show it is topologically closed".


    To that end, define for fixed x\in M the function f_x(y)=\langle x,y \rangle, \ y\in X. This function is continuous (a fundamental property of the inner product), so f_x^{-1}(\{0\})=\{y: f_x(y)=0\} is a closed set. This means

    M^{\perp}=\{y: \ \langle x,y \rangle=0, \ \forall x\in M\}=\{y: y\in  f_x^{-1}(\{0\}), \ \forall x\in M\}=\cap_{x\in M}f_x^{-1}(\{0\})<br />

    is closed as the intersection of a family of closed sets.
    Last edited by Rebesques; May 31st 2009 at 11:10 AM. Reason: comma...
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    Thank you for your replies.
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