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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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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.
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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 TB(H,K) be compact (Compact linear operator, A bounded linear operator between normed spaces) and let (sn(T))n1 be its zero-padded singular-value sequence (Absolute value and singular values of a compact operator). For n1 put an(T):=inf{TF: FB(H,K),ranF is finite-dimensional,dimFranF<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 dimFV; 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 n1, 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:HK, its singular system (ej)jJ, (fj)jJ, (sj)jJ 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)jJ(kerT) and (fj)jJranT with Tej=sjej and fj=sj1Tej, the expansion Tx=jJsjx,ejfj holds in norm, and for every n with n+1J the partial sum Tn=jnsj,ejfj satisfies TTnsn+1; moreover sn(T)=0 for all n>r when r=dimFranT<+, 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 jFsj,ejfj over a finite F has range contained in the span of the finitely many fj, hence rank at most F; for F={1,,n1} the truncation Tn1 therefore has rank <n (Finite-dimensional vector space, and its dimension dimFV; 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 dimV=dimkerF+dimranF; hence a linear map on a finite-dimensional space of dimension n with rank <n has a nonzero kernel (Rank-nullity: dimFV=nullityT+rankT, Finite-dimensional vector space, and its dimension dimFV; infinite-dimensional means having no finite basis, Kernel and image of a linear map).

[A4]

Orthonormal expansion in the span. If x=jFcjej lies in the span of finitely many members of the orthonormal family (ej), then x2=jFcj2, x,ej=cj for jF and x,ej=0 for jF; in particular, for x in the span of e1,,en the expansion of Tx reduces to the finite sum jnsjx,ejfj; 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 ainfS, and conversely ainfS 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{TF:ranF is finite-dimensional and dimFranF<n}.

1.1

Upper bound ansn. For n1 the truncation Tn1=jn1sj,ejfj (empty for n=1) has rank <n by [A2], so anTTn1. If nJ then [A1] with n1 in place of n gives TTn1sn; if nJ then r<+, n>r and Tn1=T by [A1], so an0=sn. Finally if T=0 then anT0=0=sn. In every case ansn.

A1A2A5
1.2

Lower bound snan. Let FB(H,K) have finite-dimensional range with dimFranF<n. If nJ then sn=0TF and there is nothing to prove; assume therefore nJ, so that e1,,en exist and their span V has dimension n over F by [A4]. The restriction FV:VK has rank at most dimFranF<n=dimFV, so by [A3] there is xV with x=1 and Fx=0. Writing x=jncjej with jncj2=1 by [A4], the expansion of [A1] and orthonormality of the fj give Tx2=jnsjcjfj2=jnsj2cj2sn2jncj2=sn2, the inequality because sjsn for jn by the nonincreasing order of the singular values. Hence TF(TF)x=Txsn. As F was arbitrary among the finite-rank operators with dimFranF<n, [A5] gives ansn.

A1A3A4A5algebra
2.1

Conclusion. Steps 1.1 and 1.2 give an(T)=sn(T) for every n1; 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).

step 1.1step 1.2A1A6

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