Orthocomplement is Closed Linear Subspace

Theorem
Let $\struct{ V, \innerprod \cdot \cdot }$ be an inner product space.

Let $A \subseteq V$ be a subset of $V$.

Then the orthocomplement $A^\perp$ of $A$ is a closed linear subspace of $H$.

Proof
Let $\sequence {x_n} \subset A^\perp$ be a convergent sequence.

Let $x$ be the limit of $\sequence {x_n}$.

Then, by the definition of orthocomplement:
 * $\forall n \in \N, y \in A: \innerprod {x_n} y = 0$

Passing to the limit, from Inner Product is Continuous we have:
 * $\forall y \in A: \innerprod x y = 0$

So:
 * $x \in A^\perp$

This shows that $A^\perp$ is closed.