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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?

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    $\begingroup$ $\varphi$ 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 $\{0\}$ $\endgroup$
    – Alessandro
    Apr 13, 2021 at 14:45
  • $\begingroup$ 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? $\endgroup$
    – Simeon
    Apr 13, 2021 at 14:51
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    $\begingroup$ 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) $\endgroup$
    – Alessandro
    Apr 13, 2021 at 14:53
  • $\begingroup$ 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? $\endgroup$
    – Simeon
    Apr 13, 2021 at 14:59
  • $\begingroup$ 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) $\endgroup$
    – Alessandro
    Apr 13, 2021 at 15:00

1 Answer 1

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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.

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