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

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?

• $\varphi$ is defined on $E$ (because you want the point $x\in E$ such that $\varphi(x)=0$). The reasoning is correct: If for $y$, we let $\phi_y:E\rightarrow\mathbb R$ be the map $\phi_y(x)=\langle x,y\rangle$, for any $x\in E$, then $$A^\perp=\bigcap_{y\in A}\phi^{-1}_y(\{0\})$$ and $\phi^{-1}_y(\{0\})$ is closed being the inverse image of the closed set $\{0\}$ Apr 13, 2021 at 14:45
• But this would mean that $y$ is the "fixed" vector, which would seem counterintuitive since this should apply for all $y \in A$? Also what's the deal with the intersection? Apr 13, 2021 at 14:51
• Sorry, I phrased it in the wrong way. By "for $y$, we let...." I meant "for any (or generic) $y$, we let $\phi_y$ be the map...". So what you get is a map $E\rightarrow\mathbb R$ for every $y\in A$ and $A^\perp$ is the intersection of the kernels of all these maps (which are closed sets) Apr 13, 2021 at 14:53
• Ah I see. $\phi^{-1}_y(\{0\})$ is just the preimage/kernel for a specific $y$. Taking the intersection we get the kernel for every $y$. Why it's the intersection and not union? This seems to assume that there is finitely many $y$:s? Apr 13, 2021 at 14:59
• No, the intersection of any number of closed sets is still closed, so there's no need for the $y$'s to be finite (on the other hand, is the finite union of closed sets to be closed) Apr 13, 2021 at 15:00

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