Let $Y$ be a closed subspace of the Hilbert space $X$, and let
$$Y^\perp = \{x\in X:x\perp Y\}$$
Then I want to see that
$Y^\perp$ is closed.
$X\cong Y\oplus Y^\perp$.
So for 1. I took a sequence $(x_n)\in Y^\perp $ such that $(x_n) \to x$, and I want to check that $x \in Y^\perp$, so I pick $y \in Y$ but I can't figure out how to get that the inner product $(x,y)=0$.
For 2. I wanted to use that two spaces are isomorphic iff they have the same dimension, but then I realize that $X$ could be infinite dimensinal.
So can someone help me to prove the above result please?
Thanks in advance.
Like you suggested, start with $x_n \in Y^{\perp}$ and take $y \in Y$. Since $x_n \in Y^{\perp}$, you have $\left< x_n, y \right> = 0$ for all $n \in \mathbb{N}$. If $x_n \rightarrow x$, use the continuity of the inner product to deduce that $x \in Y^{\perp}$. For the direct sum decomposition, you need to show that each vector $x \in X$ has a unique decomposition $x = y + y'$ with $y \in Y$ and $y' \in Y^{\perp}$. Choose $y$ to be the closest point to $x$ in the subspace $Y$. Show that such a $y$ indeed exists, determined uniquely by $x$ and that $x - y$ will be in $Y^{\perp}$.