Let $A \subset E$ where $E$ is an inner product space. Show that the orthogonal complement $A^{\perp}$ is closed in $E$.

Question

Let $A \subset E$ where $E$ is an inner product space. Show that the orthogonal complement $A^{\perp}$ is closed in $E$.

I have that $A^{\perp} = \{x \in E \mid \langle x, y\rangle = 0, \forall y \in A \}$. So for all fixed $x$ I have that any inner product with a vector from $A$ the result is $0$. If I denote $\varphi(y) =\langle x, y\rangle, \varphi: A \to \mathbb{R}$, then $A^{\perp} = \{x \in E \mid \varphi(y) = 0, \forall y \in A \}$, which is the preimage $\varphi^{-1}(\{0\})$ and since $\{0\}$ is closed by the fact that the metric is induced by the inner product we have that $A^{\perp}$ is closed? I'm not entirely sure I got this correctly. Is the domain for $\varphi$ correct?

Answer 1

Better define, for fixed $x \in E$, the map $\phi_x: E \to \Bbb R$ (or $\Bbb C$) by $\phi_x(y)=\langle x,y\rangle$ which is continuous.

Then $$A^\perp= \bigcap\{ \phi_x^{-1}[\{0\}]\mid x \in A\}$$

is an intersection of closed sets and hence closed.