I think this is true but I am not sure because I can't find it written in any book I have read so far and this seems to be an important result. Here is my argument:
Let $M$ be a subspace of an inner product space and let $M^{\perp}$ be it's orthogonal complement. Let x be a limit point of $M^{\perp}$. Then there is a sequence $(x_n)$ in $M^{\perp}$ such that $(x_n) \rightarrow x$. Since $\langle (x_n),y\rangle =0$ for all $y \in M$, by continuity of the inner product, $\langle x,y\rangle =0$ for all $y \in M$. This means that $x \in M^{\perp}$.
Is this proof correct?
Best Answer
Yes, the proof is correct. In fact, it is showing that $M^\perp$ is a closed subset of a Hilbert space. Since closed subsets of complete sets are complete, it follows that $M^\perp$ is complete.