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I have found a few questions similar to this one, namely If $M$ is a closed subspace of an Hilbert space $H$, then $M^{\perp\perp}=M$, but I am not assuming that $M$ is closed, so I believe this warrants a separate question. I know that I would like to use the orthogonal composition theorem, but I am not sure how to arrive there. Any hints for what proof approach I should take are appreciated. Is it as simple as a double inclusion proof?

MathStudent1324
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    If $x\perp M$, then $x\perp \overline{M}$ (closure) because of the continuity of the inner product. So $M^{\perp}=\overline{M}^{\perp}$. Then $\overline{M}=\overline{M}^{\perp\perp}=M^{\perp\perp}$. – Disintegrating By Parts May 07 '18 at 15:28
  • @Sharkos I'm not sure if this is a duplicate or not.. is there a difference between $\text{span}(A)$ and $M$? – MathStudent1324 May 07 '18 at 16:15

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