# 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 $x,y \in M \perp$

Note to be in $M \perp$ $ = = 0, \forall m \in M$

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

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

3. 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\})
$

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

4. Thank you for your replies.

,
,
,

,

,

,

,

,

,

,

,

,

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

Click on a term to search for related topics.