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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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].

1.1

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

1.2

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 $y \notin S^{⊥}$ .

A1A2A3algebra

2.1

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.

step 1.1step 1.2A4∎

Depends on

Used by

Orthonormal eigenbasis for a compact self adjoint operator Corollary
Deficiency subspaces and deficiency indices Definition
Maximal orthogonal family of cyclic reducing subspaces Lemma
Orthogonal complement of an eigenspace is invariant Lemma
Partial isometry characterizations Theorem
Spectral theorem for compact self adjoint operators Theorem
The double orthogonal complement of a subspace is its closure Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §1.3.3, p.39 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Lecture 16 (standard reference, not scraped)