Let $H$ be a Hilbert $C^*$-module over some $C^*$-algebra $A$, and consider a closed submodule $M\subseteq H$. Then it is well-known that $M$ is not necessarily orthogonally complementable in $H$. That is, we do not necessarily have (as one does when $A=\mathbb{C}$) the relation
$$H\cong M\oplus M^{\perp}.$$
As an example of this phenomenon, one can let $A = C(\mathbb{R})$, the continuous functions $\mathbb{R}\rightarrow\mathbb{C}$, let $H = A$ be the Hilbert module with the $A$-valued inner product given by $(a,b) = a^*b$, and let $M$ be the closed submodule of functions in $A$ that vanish at $0$.
However, I noticed in this case that we still have that $M^{\perp\perp}\oplus M^{\perp} = H\oplus\{0\}\cong H$.
Question: Is it always true that for a closed submodule $M$ of $H$, we have $H = M^\perp\oplus M^{\perp\perp}$?
No. In fact, let $A = C[0, 1]$ and $M = \{f \in A \mid \mathrm{supp}\, f \subset [0, 1/2]\}$. Then $M^{\perp} = \{g \mid \mathrm{supp}\, g \subset [1/2, 1]\}$, $M = M^{\perp\perp}$ and $H \ne M^{\perp\perp}\oplus M^{\perp}$.