# Thread: Annihilator of a set M in inner product space X

1. ## 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.

2. Originally Posted by frater_cp
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 $\displaystyle x,y \in M \perp$

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

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

$\displaystyle <\alpha x+\beta y,m>=<\alpha x,m> +<\beta y,m>=\alpha <x,m>+\beta <y,m>=\alpha \cdot 0 +\beta \cdot 0 =0$

3. I think the question demands "show it is topologically closed".

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

$\displaystyle M^{\perp}=\{y: \ \langle x,y \rangle=0, \ \forall x\in M\}=\{y: y\in f_x^{-1}(\{0\}), \ \forall x\in M\}=\cap_{x\in M}f_x^{-1}(\{0\})$

is closed as the intersection of a family of closed sets.

4. Thank you for your replies.

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# show that the annihilator m of a set m in an inner product space x is a closed subspace of x

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