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!!
 A: 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 $\{y_n\}_{n=1}^\infty$
A: 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.
A: 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.
