TITLE: A linear manifold in a Hilbert space is dense if and only if it has trivial orthogonal complement
QUESTION [5 upvotes]: Conway, in A Course in Functional Analysis, leaves the following corollary (2.11) to the reader.

If $\mathcal S$ is a linear manifold in $\mathcal H$, then $\mathcal
 S$ is dense in $\mathcal H$ iff $\mathcal S^\perp=(0)$.

The term linear manifold denotes a linear subspace that is not necessarily closed.

REPLY [6 votes]: As per Potato's request:
It is well know that $\mathcal{S}^{\perp\perp}=\overline{\mathcal{S}}$. So, if $
\mathcal{S}^\perp=0$, then $\overline{S}=\mathcal{S}^{\perp\perp}=\mathcal{H}$.
Conversely, if $\mathcal{S}$ is dense, and if $x\in\mathcal{S}^\perp$, then $f(v):=\langle v,x\rangle$ is a continuous function vanishing on the dense set $\mathcal{S}$, and thus must be identically zero. Thus, $x\in\mathcal{H}^\perp=0$.