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 $\displaystyle \alpha, \beta \in \mathbb{R}$
$\displaystyle <\alpha x+\beta y,m>=<\alpha x,m> +<\beta y,m>=\alpha <x,m>+\beta <y,m>=\alpha \cdot 0 +\beta \cdot 0 =0$
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.