I just have a quick question. Suppose there is $Z$, a Hilbert space, with $A$ and $B$ closed linear subspaces. If $(a,b)=0$ for all $a \in A$ and $b \in B$, I know that $A+B$ is also closed. I don't know how to prove it though.
I am just not sure where to start. Should I try to show the complement is open, or should I do something with sequences in $M+N$? I would really appreciate a little hint.
Note that every element of the space $A+B$ has a unique representation $a+b$ where $a\in A,b\in B$. Now consider any sequence of points in $A+B$, it can be written as $\{a_n+b_n\}_{n=1,2,3,\ldots}$. Suppose that $$\lim(a_n+b_n)=x\in Z.$$ It is enough to show that $x\in A+B$. Let $\pi_A$ and $\pi_B$ denote orthogonal projections $Z\rightarrow A$ and $Z\rightarrow B$ respectively. Note that $\pi_A(a_n+b_n)=a_n$ and $\pi_b(a_n+b_n)=b_n$. Hence $\lim\pi_A(a_n+b_n)=\lim a_n=\pi_A(x)$ and $\lim\pi_B(a_n+b_n)=\lim b_n=\pi_B(x)$ (since projections are continuous). Therefore $\lim(a_n+b_n)=x=\pi_A(x)+\pi_B(x)\in A+B$. QED