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

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The double orthogonal complement of a subspace is its closure

Statement

Assume the Axiom of Countable Choice. Let $M$ be a linear subspace of a real or complex Hilbert space $H$ . Then

$M^{⊥ ⊥} = \overline{M},$

where $\overline{M}$ is the norm closure of $M$ and $M^{⊥ ⊥} = (M^{⊥})^{⊥}$ .

Facts & Assumptions

[A1]

$S^{⊥}$ is a linear subspace, $S \subseteq S^{⊥ ⊥}$ , and $S \subseteq T$ implies $T^{⊥} \subseteq S^{⊥}$ (Orthogonality and the orthogonal complement).

[A2]

$S^{⊥}$ is closed for every subset $S$ , and the closure of a set is the smallest closed superset, so $\overline{M}$ is contained in every closed set containing $M$ . A point lies in $\overline{M}$ exactly when every norm ball about it meets $M$ (Orthogonal complements are closed, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[A3]

$M^{⊥ ⊥}$ is a linear subspace and $M^{⊥ ⊥} \cap M^{⊥} = {0}$ (Orthogonality and the orthogonal complement).

[A4]

The inner-product norm is homogeneous and satisfies the triangle inequality (The induced length is a norm).

[A5]

Every closed linear subspace of $H$ splits $H$ as $H = N \oplus N^{⊥}$ (Orthogonal decomposition by a closed subspace).

[A6]

Countable Choice is the hypothesis under which the decomposition is available (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: Countable Choice, a Hilbert space $H$ and a linear subspace $M \subseteq H$ .

1.1A1A2A6

Every $m \in M$ is orthogonal to every element of $M^{⊥}$ , so $M \subseteq M^{⊥ ⊥}$ ; since $M^{⊥ ⊥}$ is closed by [A2], and $\overline{M}$ is the smallest closed superset of $M$ , we get $\overline{M} \subseteq M^{⊥ ⊥}$ .

1.2A2A4

The closure $N = \overline{M}$ is a linear subspace. It contains $0 \in M$ . If $u, v \in N$ and $r > 0$ , choose $u^{'}, v^{'} \in M$ with $∥ u - u^{'} ∥ < r / 2$ and $∥ v - v^{'} ∥ < r / 2$ ; then $u^{'} + v^{'} \in M$ and $∥ (u + v) - (u^{'} + v^{'}) ∥ < r$ , so every ball about $u + v$ meets $M$ and $u + v \in N$ . If $a = 0$ , then $a u = 0 \in N$ ; if $a \neq 0$ , for every $r > 0$ choose $u^{'} \in M$ with $∥ u - u^{'} ∥ < r / ∣ a ∣$ , and then $a u^{'} \in M$ and $∥ a u - a u^{'} ∥ < r$ , so $a u \in N$ .

2.1step 1.1step 1.2A1A3A5

Let $x \in M^{⊥ ⊥}$ and decompose $x = n + z$ with $n \in N$ and $z \in N^{⊥}$ ; then $z = x - n$ lies in $M^{⊥ ⊥}$ because both $x$ and $n \in N \subseteq M^{⊥ ⊥}$ do, while $z \in N^{⊥} \subseteq M^{⊥}$ because $M \subseteq N$ ; hence $z \in M^{⊥} \cap M^{⊥ ⊥} = {0}$ , so $z = 0$ and $x = n \in \overline{M}$ .

3.1step 1.1step 2.1∎

Therefore $M^{⊥ ⊥} \subseteq \overline{M}$ by step 2.1, and the reverse inclusion is step 1.1, so $M^{⊥ ⊥} = \overline{M}$ for every linear subspace $M$ .

The double orthogonal complement of a subspace is its closure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources