1. The problem asks us to prove that for a set $C$, the double orthogonal complement $C^{\perp\perp}$ equals $C$. 2. Recall the definition: For a set $C$ in an inner product space, the orthogonal complement $C^\perp$ is the set of all vectors orthogonal to every vector in $C$. 3. The double orthogonal complement $C^{\perp\perp}$ is the orthogonal complement of $C^\perp$, i.e., the set of all vectors orthogonal to every vector in $C^\perp$. 4. We want to show $C^{\perp\perp} = C$. This is a standard result in Hilbert spaces and finite-dimensional inner product spaces. 5. First, show $C \subseteq C^{\perp\perp}$: Every vector in $C$ is orthogonal to every vector in $C^\perp$ by definition, so $C$ is contained in $C^{\perp\perp}$. 6. Next, show $C^{\perp\perp} \subseteq C$: If a vector $x$ is in $C^{\perp\perp}$, then $x$ is orthogonal to every vector orthogonal to $C$. By the properties of inner product spaces, this implies $x$ lies in the closure of the span of $C$. If $C$ is closed, then $x \in C$. 7. Therefore, $C^{\perp\perp} = C$ when $C$ is a closed subspace. Final answer: $$C^{\perp\perp} = C$$