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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Singular values equal approximation numbers

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Let H and K be Hilbert spaces over the same field F∈{R,C}, let T∈B(H,K) be compact (Compact linear operator, A bounded linear operator between normed spaces) and let (sn(T))n≥1 be its zero-padded singular-value sequence (Absolute value and singular values of a compact operator). For n≥1 put an(T):=inf⁡{∥T−F∥: F∈B(H,K),ran⁡F is finite-dimensional,dim⁡Fran⁡F<n} (Greatest lower bound (infimum), The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis). The displayed set is nonempty because it contains F=0, and it is bounded below by 0; its infimum therefore exists by the real infimum property (Every nonempty set bounded below has an infimum). Then an(T)=sn(T)for every n≥1, including the zero-padded case: if T has finite rank r and n>r then both numbers are 0, and if T=0 both are 0 for every n.

Facts & Assumptions

Given: Countable Choice, a compact T:H→K, its singular system (ej)j∈J, (fj)j∈J, (sj)j∈J from the singular-value decomposition, and the numbers an(T).

[A1]

Singular-value decomposition. With the index set J of the positive singular values with multiplicity, there are orthonormal systems (ej)j∈J⊆(ker⁡T)⊥ and (fj)j∈J⊆ran⁡T‾ with ∣T∣ej=sjej and fj=sj−1Tej, the expansion Tx=∑j∈Jsj⟨x,ej⟩fj holds in norm, and for every n with n+1∈J the partial sum Tn=∑j≤nsj⟨⋅,ej⟩fj satisfies ∥T−Tn∥≤sn+1; moreover sn(T)=0 for all n>r when r=dim⁡Fran⁡T<+∞, and J={1,…,r} in that case (Singular value decomposition for compact operators, Absolute value and singular values of a compact operator).

[A2]

Ranks of truncations. A finite sum ∑j∈Fsj⟨⋅,ej⟩fj over a finite F has range contained in the span of the finitely many fj, hence rank at most ∣F∣; for F={1,…,n−1} the truncation Tn−1 therefore has rank <n (Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis, Kernel and image of a linear map).

[A3]

Rank–nullity in finite dimensions. A linear map of a finite-dimensional space V satisfies dim⁡V=dim⁡ker⁡F+dim⁡ran⁡F; hence a linear map on a finite-dimensional space of dimension n with rank <n has a nonzero kernel (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis, Kernel and image of a linear map).

[A4]

Orthonormal expansion in the span. If x=∑j∈Fcjej lies in the span of finitely many members of the orthonormal family (ej), then ∥x∥2=∑j∈F∣cj∣2, ⟨x,ej⟩=cj for j∈F and ⟨x,ej⟩=0 for j∉F; in particular, for x in the span of e1,…,en the expansion of Tx reduces to the finite sum ∑j≤nsj⟨x,ej⟩fj; the finite Bessel inequality bounds partial sums of coefficients (Orthonormal families, complete orthonormal systems and Hilbert bases, The finite Bessel inequality and best approximation by a finite orthonormal family, Real and complex inner-product spaces and their induced length).

[A5]

Infimum. Every nonempty lower-bounded subset of R has an infimum (Every nonempty set bounded below has an infimum); by the defining greatest-lower-bound property, every lower bound a of S satisfies a≤inf⁡S, and conversely a≤inf⁡S makes a a lower bound of S (Greatest lower bound (infimum)).

[A6]

Countable Choice is the standing hypothesis of this pair's Hilbert-space interface (The Axiom of Countable Choice (ACω)).

Proof

technique · direct

Given: Countable Choice, the compact T, its singular system and the numbers an(T)=inf⁡{∥T−F∥:ran⁡F is finite-dimensional and dim⁡Fran⁡F<n}.

1.1A1A2A5

Upper bound an≤sn. For n≥1 the truncation Tn−1=∑j≤n−1sj⟨⋅,ej⟩fj (empty for n=1) has rank <n by [A2], so an≤∥T−Tn−1∥. If n∈J then [A1] with n−1 in place of n gives ∥T−Tn−1∥≤sn; if n∉J then r<+∞, n>r and Tn−1=T by [A1], so an≤0=sn. Finally if T=0 then an≤∥T−0∥=0=sn. In every case an≤sn.

1.2A1A3A4A5algebra

Lower bound sn≤an. Let F∈B(H,K) have finite-dimensional range with dim⁡Fran⁡F<n. If n∉J then sn=0≤∥T−F∥ and there is nothing to prove; assume therefore n∈J, so that e1,…,en exist and their span V has dimension n over F by [A4]. The restriction F∣V:V→K has rank at most dim⁡Fran⁡F<n=dim⁡FV, so by [A3] there is x∈V with ∥x∥=1 and Fx=0. Writing x=∑j≤ncjej with ∑j≤n∣cj∣2=1 by [A4], the expansion of [A1] and orthonormality of the fj give ∥Tx∥2=∥∑j≤nsjcjfj∥2=∑j≤nsj2∣cj∣2≥sn2∑j≤n∣cj∣2=sn2, the inequality because sj≥sn for j≤n by the nonincreasing order of the singular values. Hence ∥T−F∥≥∥(T−F)x∥=∥Tx∥≥sn. As F was arbitrary among the finite-rank operators with dim⁡Fran⁡F<n, [A5] gives an≥sn.

2.1step 1.1step 1.2A1A6∎

Conclusion. Steps 1.1 and 1.2 give an(T)=sn(T) for every n≥1; in the finite-rank case with n>r both sides are 0 by [A1] and [step 1.1], and for T=0 the equality reads an(0)=0=sn(0).

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Dependency tree · two levels

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Sources