Let $V$ be a Hilbert space and $U \subseteq V$. Then $U^\perp = \{\mathbf{v} \in V|\forall \mathbf{u} \in U, \langle \mathbf{u}, \mathbf{v} \rangle = 0 \}$.
My question is, how do you express $\left(U^{\perp}\right)^{\perp}$? I attempted to simply chain two definitions together. $\left(U^{\perp}\right)^{\perp} = \{\mathbf{w} \in V|\forall \mathbf{v} \in V, \forall \mathbf{u} \in U, \langle \mathbf{u}, \mathbf{v} \rangle = 0| \langle \mathbf{v}, \mathbf{w} \rangle = 0\}$, but I don't think that is valid.
It's hard to say whether your expression is valid. However, it is certainly unclear since you include more than one $\mid$ in your set definition. I would write something like this: $$ \newcommand{\ip}[1]{\langle#1 \rangle} (U^\perp)^\perp = \{w \in V: v \in U^\perp \implies \ip{v,w} = 0\}\\ = \{w \in V: (\forall u \in U, \ip{v,u} = 0) \implies \ip{v,w} = 0\} $$ That being said, even this isn't the simplest expression for $(U^\perp)^\perp$. We can show that, in general, $(U^\perp)^\perp = \overline{U}$.