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!!

  • 2
    $\begingroup$ Yes, your proof is correct! $\endgroup$ – Janko Bracic Feb 21 '15 at 15:58
  • 2
    $\begingroup$ yup thats a good proof :) $\endgroup$ – Aerinmund Fagelson Feb 21 '15 at 16:04
  • $\begingroup$ Just a question, is $A$ closed? or arbitrary? $\endgroup$ – sleeve chen Apr 18 '17 at 9:59
  • $\begingroup$ 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 $x\in A$". $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.