TITLE: Hilbert space, dense, orthogonal complement.
QUESTION [5 upvotes]: Suppose $H$ is a Hilbert space, $A$ and $B$ are two subspaces. $A$ is closed and $B$ is dense.
If $A^\perp \cap B=\{0\}$, or in other words, $\forall b\in B$, the projection to $A$ is not $0$, can we conclude that $A$ is the whole space?
Thanks,

REPLY [7 votes]: If $B = H$, we can of course conclude it. Otherwise, choose an $x \in H \setminus B$, and consider $A = x^\perp$. Then $A^\perp = \operatorname{span} \{x\}$, hence $A^\perp \cap B = \{0\}$, but $A \neq H$.