Orthocomplement is Closed Linear Subspace

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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.

$\blacksquare$


Sources