To show that orthogonal complement of a set A is closed.

To show that orthogonal complement of a set A is closed.

My try: I first show that the inner product is a continuous map. Let $X$ be an inner product space. For all $x_1,x_2,y_1,y_2 \in X$, by Cauchy-Schwarz inequality we get, $$|\langle x_1,y_1\rangle - \langle x_2,y_2\rangle| = |\langle x_1- x_2,y_1\rangle + \langle x_2, y_1-y_2\rangle|$$ $$\leq \|x_1- x_2\|\cdot\|y_1\| +\|x_2\|\cdot\| y_1-y_2\|$$

This implies continuity of inner products.

Let $A \subset X$ and $y \in A^\perp$. To show that $A^\perp$ is closed, we have to show that if $(y_n)$ is convergent sequence in $A^\perp$, then the limit $y$ also belong to $A^\perp$.

Let $x \in A$, then using that inner product is a continuous map, $$\langle x,y\rangle = \langle x, \lim_{n\to \infty} (y_n)\rangle = \lim_{n\to \infty} \langle x, y_n\rangle = 0.$$

Since $\langle x, y_n\rangle = 0$ for all $x \in A$ and $y_n \in A^\perp$. Hence $y \in A^\perp$.

Is the approach\the proof correct??

Thank You!!

• Yes, your proof is correct! – Janko Bracic Feb 21 '15 at 15:58
• yup thats a good proof :) – Aerinmund Fagelson Feb 21 '15 at 16:04
• Just a question, is $A$ closed? or arbitrary? – sleeve chen Apr 18 '17 at 9:59
• The result holds for arbitrary $A\subseteq X$, including the somewhat peculiar cases $A=\varnothing$ and $A=X$ (it's a good exercise to check what closed subspaces $\varnothing^\perp$ and $X^\perp$ actually correspond to). But, if OP wants their proof to work for $A=\varnothing$ they need to be a bit careful and rephrase the sentence that starts with "let $x\in A$" to something like "for any $x\in A$". – Oskar Henriksson Nov 22 '17 at 4:58

I really like your proof, so formalizing it we have:

Let be $\{y_n\}_{n=1}^\infty \in A^\perp$ s.t. $y_n \to y$ and let be $x \in A$.

We now want to show that $y\in A^\perp$.

From the inner product's continuity we have:

$\forall \epsilon>0\ ,\exists\ \delta>0$ such that:

$|\langle x, y_n-y\rangle|<\epsilon$, if $\parallel y_n-y\parallel<\delta$ **

we shall now see that $\langle x, y_n\rangle = 0\ \forall n \in \mathbb N$

then $|\langle x, y_n\rangle - \langle x, y\rangle| = |\langle x, y\rangle|<\epsilon$ , which implies $\langle x, y\rangle = 0$

this means $y\in A^\perp$ q.e.d.

** Using the norm induced by the inner product, we may also note the existence of $\delta$ is guaranteed from convergence of $\{y_n\}_{n=1}^\infty$

Let $$\{y_n\}_{n=1}^\infty \in A^\perp$$ s.t. $$y_n \to y.$$

Then $$\forall x\in A,$$ we have $$\langle y , x\rangle=\lim_{n\to\infty} \langle y_n , x\rangle=\lim_{n\to\infty}0=0,$$

where the first equality holds because of the (norm) convergence of $$y_n$$ to $$y$$ and of the Cauchy Schwartz inequality.

So $$y\in A^\perp$$ as desired.

Let's denote the orthogonal complement of $$A$$ by $$A^{\perp}$$. Also, we denote the scalar product $$\langle \cdot, y \rangle : V \rightarrow \mathbb{F}$$ as the function $$\varphi_y$$. So, all the orthogonal elements corresponding to $$y$$ is nothing but $$Ker \; \varphi_y$$. Hence,

$$A^{\perp} = \cap_{y \in A} \varphi_y$$

Since $$Ker \; \varphi_y$$ is closed subset of $$V$$ (why?) and arbitrary intersection of closed set is again a closed set, we have the result.