How do I prove that if $A$ and $B$ are complementary then $A^\perp$ and $B^\perp$ are also complementary in $\mathcal{H}$ Hilbert space? I have the following problem: I know that $A$ and $B$ are two closed complementary subspaces in an Hilbert space $\mathcal{H}$, i.e. $\mathcal{H}=A\oplus B$ (where $\oplus$ is the direct sum of two closed subspaces). I have to prove that $A^\perp$ and $B^\perp$ are also complementary. 
For now, I only managed to prove that $\overline{A^\perp\oplus B^\perp}=\mathcal{H}$ using the identity $(R\oplus S)^{\perp}=R^{\perp}\cap S^{\perp}$ and plugging in $R=A^{\perp}, S=B^\perp$ and taking into account that $A$ and $B$ are closed to say $(A^\perp)^\perp=A$, $(B^\perp)^\perp=B$. 
I know that in general the direct sum of two closed subspaces is not closed, so I cannot conclude that $\overline{A^\perp\oplus B^\perp}=A^\perp\oplus B^\perp$. If someone can give me an hint on how to solve this it would be greatly appreciated.
 A: To begin with, two subspaces $V$ and $W$ do not always have an (internal) direct sum, they do have a sum $V+W$ which is called direct when $V \cap W = \{0\}$. So we have to prove that $A^\bot \cap B^\bot = \{0\}$ and $A^\bot + B^\bot = \mathcal H$. The first thing is easy, so I leave it "to the reader". The second is less trivial; it can be done like this: It may help to set $S = A^\bot + B^\bot$ for better understanding of what follows. Now we want to prove that $S = \mathcal H$; for this we show that $S^\bot = \{0\}$. Indeed, suppose that $x$ is an element of this orthogonal subspace. This means that $x$ is orthogonal to $A^\bot + B^\bot$, which implies that it is orthogonal to $A^\bot$ and $B^\bot$, in other words it belongs to $A^{\bot\bot}$ and $B^{\bot\bot}$, which are equal respectively to $A$ and $B$ because these are closed subspaces (an elementary property of orthogonal subspaces in a Hilbert space). So we get that $x \in A \cap B = \{0\}$, i.e. $x = 0$, proving our claim. Therefore $S = S+S^\bot = \mathcal H$ and we are done.
