LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Orthogonal complements are closed

Statement

For every subset $S$ of a real or complex inner-product space $V$ , the orthogonal complement $S^{⊥}$ is a closed linear subspace of $V$ for the induced norm topology.

Facts & Assumptions

[A1]

$S^{⊥} = {v \in V : ⟨ v, s ⟩ = 0 for all s \in S}$ is a linear subspace and orthogonality is symmetric (Orthogonality and the orthogonal complement).

[A2]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A3]

The induced length is a norm, so $∥ v ∥ \geq 0$ with $∥ v ∥ = 0$ exactly for $v = 0$ (The induced length is a norm).

[A4]

In the metric topology a set is open exactly when every point of it has a ball around it inside the set, and $B (x, r) = {y : d (x, y) < r}$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

Given: A subset $S$ of a real or complex inner-product space $V$ , with $S^{⊥}$ as in [A1].

1.1A1

By [A1] the set $S^{⊥}$ is a linear subspace of $V$ , which is the first assertion.

1.2A1A2A3algebra

Let $x \notin S^{⊥}$ ; then some $s \in S$ has $⟨ x, s ⟩ \neq 0$ , so $s \neq 0$ and $r = ∣ ⟨ x, s ⟩ ∣ / (2 ∥ s ∥) > 0$ ; if $y \in V$ satisfies $∥ y - x ∥ < r$ , then $∣ ⟨ y, s ⟩ ∣ \geq ∣ ⟨ x, s ⟩ ∣ - ∣ ⟨ x - y, s ⟩ ∣ > ∣ ⟨ x, s ⟩ ∣ - r ∥ s ∥ = ∣ ⟨ x, s ⟩ ∣ / 2 > 0$ by Cauchy–Schwarz, so $y \notin S^{⊥}$ .

2.1step 1.1step 1.2A4∎

Thus every point outside $S^{⊥}$ has a ball around it that misses $S^{⊥}$ , so $V ∖ S^{⊥}$ is open and $S^{⊥}$ is closed in the metric topology; with step 1.1 this proves that $S^{⊥}$ is a closed linear subspace.

Orthogonal complements are closed

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources