Let $X$ be a vector space, with a pseudo inner product space (means that it doesnt have to satisfy that $\langle u,u\rangle =0 \Leftrightarrow u=0$). Let $Y$ be a subspace of $X$. Show that the orthogonal complement of $Y$ defined as:
$$Y^{\perp}=\{x\in X:\langle x,z\rangle=0 \forall z\in Y\}$$
Show that $Y^{\perp}$ is closed.
What i've tried/done so far:
Let $(u_n)$ be a sequence of vectors from $Y^{\perp}$, which converges to a vector $u\in X$. Now if I can show that $u$ is in fact a vector in $Y^{\perp}$, then the set most the closed. So basically I want to show that $\langle u,z\rangle=0\forall z\in Y$.
I have a Lemma that states that for vectors $u,q,u_1,q_1,u_2,q_2,… \in X$ then
$$\lim_{n\rightarrow \infty}\|u_n-u\|=0=\lim_{n\rightarrow \infty}\|q_n-q\|$$
and
$$\lim_{n\rightarrow \infty}\langle u_n,q_n\rangle =\langle u,q\rangle$$
But I havent come any further, any help is much appreciated.
Best Answer
For $x \in X$, define $x_\flat: X\to \Bbb R$ by $x_\flat(y)\doteq \langle x,y\rangle$. You can check that $x_\flat$ is continuous. So $x_\flat^{-1}(0)$ is closed. And $Y^\perp =\bigcap_{x\in Y} x_\flat^{-1}(0)$ is an intersection of closed sets.