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 $M \subseteq H$ .

1.1

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 $\overline{M} \subseteq M^{⊥ ⊥}$ .

A1A2A6

1.2

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 $a u \in N$ .

A2A4

2.1

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 $x = n \in \overline{M}$ .

step 1.1step 1.2A1A3A5

3.1

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 $M$ .

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

40 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, Lemma 5.37, p.238 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Theorem 181 (standard reference, not scraped)