How to prove $ A^{\perp} $ is a closed linear subspace? Suppose $ X $ is an inner product space and $ A\subseteq X $. I need to prove that $ A^{\perp} $ is a closed linear subspace of $ X $. Can anyone give me a idea?
 A: Let $ x,y\in A^{\perp} $, $ a\in A $ and let $ \alpha ,\beta $ be two field elements of $ X $. Then $$ \langle \alpha x+\beta y,a\rangle =\alpha \langle x,a\rangle +\beta \langle y,a\rangle =0 .$$ Therefore $ \alpha x+\beta y\in A^{\perp} $  and hence $ A^{\perp} $ is a liner subspace. 
To show $ A^{\perp} $ is closed, let $ (x_{n}) $ be a sequence in $ A^{\perp} $ such that $ (x_{n}) $ converges to $ x $. Observe that for all $ a\in A $, $$ 0=\langle 0,a\rangle =\lim_{n\rightarrow \infty}\langle x-x_{n},a\rangle =\langle x,a\rangle -\lim_{n\rightarrow \infty}\langle x_{n},a\rangle =\langle x,a\rangle .$$ Therefore $ x\in A^{\perp} $ and hence $ A^{\perp} $ is closed.
A: Let be a point $a\in X$:
$$\{a\}^{\perp}=\{x\in X\,\vert\,\langle a,x\rangle=0\}$$
is closed (inverse image of the closed set $\{0\}$ by a continuous function). And
$$A^{\perp}=\bigcap_{a\in A}\{a\}^{\perp}$$
is an intersection of closed sets.
A: take a sequence $x_n$ in $ A^{\perp}$. such that $x_n\rightarrow x$ 
see that 
as $\langle x_n,a\rangle=0\;\forall a\in A$ $\implies$ $\langle x,a\rangle=0\;\forall a\in A$ $\implies$ $x\in A^{\perp}$ and $A^{\perp}$ is closed
And you can verify $A^{\perp}$ is subspace same way
A: Notation: We use $Y\subseteq X$ instead.
Show that $Y^\perp$ is a closed linear subspace of $X'$.
Let $\alpha,\beta\in\mathbb{C}$, $l_1,l_2\in Y^\perp$, $y\in Y$.
$(\alpha l_1+\beta l_2)(y)=\alpha l_1(y)+\beta l_2(y)=0$. Therefore $Y^\perp$ is linear.
Let $(l_n)$ be a sequence in $Y^\perp$ converging to $l\in X'$, i.e.\ $\|l_n-l\|\to0$. There exists $N$ such that for all $n\geq N$, $\|l_n-l\|=\sup_{x\leq 1}|(l_n-l)(x)|<\epsilon$.
Let $y\in Y$. For $n\geq N$, we have
\begin{align*}
|l(y)|&=\|y\||l(\frac{y}{\|y\|})|\\
&\leq\|y\||(l-l_n)(\frac{y}{\|y\|})|+\|y\||l_n(\frac{y}{\|y\|})|\\
&<\|y\|\epsilon.
\end{align*}
Since $\epsilon$ is arbitrary, $l(y)=0$ for all $y\in Y$. Thus $l\in Y^\perp$.
A: Let $\psi_a: X\to \mathbb{C}$ be defined by $b\mapsto \langle a,b\rangle$ then $A^\perp = \cap_{a\in A} \psi^{-1}_a (0)$ so it suffices to show that $\psi^{-1}_a (0)$ is closed for any $a$ (equivalently that $X \backslash \psi^{-1}_a (0)$ is open).
$$X \backslash \psi^{-1}_a (0) = \{ b \in X | \langle a,b\rangle \in \mathbb{C}\backslash \{0\} \} = \psi^{-1}_a (\mathbb{C}\backslash \{0\})$$ is open since it's the continuous preimage of an open set (singletons are closed).
