Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
How statement and proof provenance work

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.

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 (xn)nN be a sequence in H whose range {xn:nN} is dense in H (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets), and let

L0:=,Ln+1:=Ln{vnvn}  if  vn:=xneLnxn,ee0,Ln+1:=Ln  otherwise.

Then L:=nNLn 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 xH, then v:=xeEx,ee is orthogonal to every element of E; if v0 then v/v has norm 1 and E{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:AA and initial state a0A, recursion on N produces the unique sequence with an+1=f(an) (The recursion theorem).

[A3]

The span of a finite orthonormal set is a linear subspace, and x=P+v with P=eEx,eespanE and vspan(E{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; v2=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).

[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 (xn)nN in the Hilbert space H and the Gram–Schmidt sets Ln defined by the displayed recursion.

1.1

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×O. Define f(n,E)=(n+1,Φn(E)), where Φn(E) is the displayed update computed from xn 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,)A. Recursion from (0,) gives states (n,Ln) and hence the required sets Ln. By induction on n, each Ln is a finite orthonormal set with LnLn+1. Indeed L0= is orthonormal, and if Ln is finite and orthonormal then vn is orthogonal to every element of Ln; either vn=0 and Ln+1=Ln, or vn0 and Ln+1=Ln{vn/vn} is orthonormal.

A1A2
2.1

L=nN(Ln+1Ln) and each difference Ln+1Ln has at most one element; hence the map assigning to every n with Ln+1Ln its unique element is a bijection onto L from the subset J={n:Ln+1Ln} 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=.

step 1.1A6
2.2

By induction on n one has spanLn=span{x0,,xn1}: for n=0 both sides are {0}, and if xn=P+vn with PspanLn and vnspan(Ln{vn})=spanLn+1, then spanLn+1=span(Ln{xn})=span{x0,,xn}, the case vn=0 included.

step 1.1A3
3.1

Every xn lies in spanLn+1spanL, 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 Ln, by taking the larger of their finite appearance stages; hence L is orthonormal by step 1.1. Its closed linear span is therefore H.

step 1.1step 2.1step 2.2A4A5
4.1

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.

step 2.1step 3.1

Depends on

Used by

Dependency tree · two levels

55 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