TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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A Hilbert space with a dense sequence has a finite or countable orthonormal basis

Statement

Let $H$ be a real or complex Hilbert space, let $(x_{n})_{n \in N}$ be a sequence in $H$ whose range ${x_{n} : n \in N}$ is dense in $H$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets), and let

$L_{0} : = \emptyset, L_{n + 1} : = L_{n} \cup {\frac{v_{n}}{∥ v_{n} ∥}} if v_{n} : = x_{n} - \sum_{e \in L_{n}} ⟨ x_{n}, e ⟩ e \neq 0, L_{n + 1} : = L_{n} otherwise .$

Then $L : = ⋃_{n \in N} L_{n}$ is a finite or countably infinite orthonormal set whose closed linear span is $H$ ; that is, $L$ is an orthonormal basis of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), and it is obtained from the given sequence by Gram–Schmidt elimination. The enumeration of $L$ is the canonical one by the stage at which an element appears, and no choice principle is used.

Facts & Assumptions

[A1]

If $E$ is a finite orthonormal set in $H$ and $x \in H$ , then $v : = x - \sum_{e \in E} ⟨ x, e ⟩ e$ is orthogonal to every element of $E$ ; if $v \neq 0$ then $v / ∥ v ∥$ has norm $1$ and $E \cup {v / ∥ v ∥}$ is orthonormal (The finite Bessel inequality and best approximation by a finite orthonormal family, The induced length is a norm). Finite vector sums are independent of an enumeration because vector addition is a commutative monoid; the empty sum is zero (A finite sum in a commutative monoid indexed by an arbitrary finite set, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[A2]

For a fixed function $f : A \to A$ and initial state $a_{0} \in A$ , recursion on $N$ produces the unique sequence with $a_{n + 1} = f (a_{n})$ (The recursion theorem).

[A3]

The span of a finite orthonormal set is a linear subspace, and $x = P + v$ with $P = \sum_{e \in E} ⟨ x, e ⟩ e \in span E$ and $v \in span (E \cup {v})$ (Linear subspace of a vector space, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[A4]

An orthonormal set is complete when its closed linear span is $H$ ; $∥ v ∥^{2} = ⟨ v, v ⟩$ and $∥ v ∥ = 1$ says $⟨ v, v ⟩ = 1$ (Orthonormal families, complete orthonormal systems and Hilbert bases, Real and complex inner-product spaces and their induced length).

[A5]

The range of $(x_{n})$ is dense in $H$ : its closure is $H$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[A6]

Every subset of $N$ is at most countable, without choice, and an infinite subset has its canonical increasing enumeration (Every subset of an at most countable set is at most countable). A set in bijection with a finite or countably infinite set is itself finite or countably infinite (Finite, countably infinite, countable, uncountable).

Proof

technique · direct

Given: A dense sequence $(x_{n})_{n \in N}$ in the Hilbert space $H$ and the Gram–Schmidt sets $L_{n}$ defined by the displayed recursion.

1.1A1A2

To put the stage-dependent rule into the fixed-function form of [A2], let $O$ be the set of finite orthonormal subsets of $H$ , a subset of $P (H)$ , and use the state space $A = N \times O$ . Define $f (n, E) = (n + 1, Φ_{n} (E))$ , where $Φ_{n} (E)$ is the displayed update computed from $x_{n}$ and the finite set $E$ . The sum exists by [A1]; in the nonzero residual branch its norm is positive and normalization is defined. The update is again finite and orthonormal by [A1], and in the zero branch it is $E$ . Thus $f$ is a total self-map of $A$ , and $(0, \emptyset) \in A$ . Recursion from $(0, \emptyset)$ gives states $(n, L_{n})$ and hence the required sets $L_{n}$ . By induction on $n$ , each $L_{n}$ is a finite orthonormal set with $L_{n} \subseteq L_{n + 1}$ . Indeed $L_{0} = \emptyset$ is orthonormal, and if $L_{n}$ is finite and orthonormal then $v_{n}$ is orthogonal to every element of $L_{n}$ ; either $v_{n} = 0$ and $L_{n + 1} = L_{n}$ , or $v_{n} \neq 0$ and $L_{n + 1} = L_{n} \cup {v_{n} / ∥ v_{n} ∥}$ is orthonormal.

2.1step 1.1A6

$L = ⋃_{n \in N} (L_{n + 1} ∖ L_{n})$ and each difference $L_{n + 1} ∖ L_{n}$ has at most one element; hence the map assigning to every $n$ with $L_{n + 1} ∖ L_{n} \neq \emptyset$ its unique element is a bijection onto $L$ from the subset $J = {n : L_{n + 1} ∖ L_{n} \neq \emptyset}$ of $N$ : surjectivity follows from the union, and outputs at distinct stages are distinct because the sets are increasing and only new elements enter a difference. By [A6], $J$ and hence $L$ are finite or countably infinite. Ordering the stages increasingly gives the canonical enumeration, including the empty one when $J = \emptyset$ .

2.2step 1.1A3

By induction on $n$ one has $span L_{n} = span {x_{0}, \dots, x_{n - 1}}$ : for $n = 0$ both sides are ${0}$ , and if $x_{n} = P + v_{n}$ with $P \in span L_{n}$ and $v_{n} \in span (L_{n} \cup {v_{n}}) = span L_{n + 1}$ , then $span L_{n + 1} = span (L_{n} \cup {x_{n}}) = span {x_{0}, \dots, x_{n}}$ , the case $v_{n} = 0$ included.

3.1step 1.1step 2.1step 2.2A4A5

Every $x_{n}$ lies in $span L_{n + 1} \subseteq span L$ , so the span of $L$ contains the whole dense range of the sequence; hence its closure contains the closure of that range, which is $H$ , while it is itself contained in $H$ . Furthermore, any two elements of $L$ belong to one common $L_{n}$ , by taking the larger of their finite appearance stages; hence $L$ is orthonormal by step 1.1. Its closed linear span is therefore $H$ .

4.1step 2.1step 3.1∎

By steps 3.1 and 2.1 the set $L$ is a finite or countably infinite orthonormal set with closed linear span $H$ , that is, an orthonormal basis of $H$ obtained by Gram–Schmidt elimination from the given dense sequence, canonically enumerated by the stages at which its elements appear.

A Hilbert space with a dense sequence has a finite or countable orthonormal basis

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources