Alphabeta Math
CorollaryStatement: 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.

Spectrum of a compact operator is countable with only zero as possible accumulation

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let X be a complex Banach space, let K:XX be a compact operator (Compact linear operator) and let σ(K) be its spectrum (Spectrum and resolvent of a bounded operator). Then:

  1. σ(K) is at most countable (Finite, countably infinite, countable, uncountable);
  2. for every λC with λ0 there is a real r>0 with B(λ,r)σ(K){λ} (Open ball, closed ball and sphere in a metric space).

In particular the only point of C that can be an accumulation point of σ(K) (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) is 0.

Facts & Assumptions

[A1]

Under AC, for every real ε>0 the set Sε:={λσ(K):λε} is finite (Riesz schauder spectrum of a compact operator).

[A2]

For every real ε>0 there is a natural n1 with 1/n<ε (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[A4]

In C the balls are those of the metric d(z,w)=zw (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, Open ball, closed ball and sphere in a metric space); a point λ is an accumulation point of a set A when every punctured ball B(λ,r){λ} meets A (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Proof

technique · direct

Given: AC, a complex Banach space X, a compact K:XX, and the sets S1/n={λσ(K):λ1/n}, n1.

1.1

Every subspace of σ(K) of the form S1/n is finite by [A1]; and every μσ(K) with μ0 lies in some S1/n, because μ>0 and [A2] gives n with 1/n<μ.

A1A2
1.2

The set U:={0}n1S1/n is at most countable: it is a countable union of finite sets, and [A3] applies.

A1A3
2.1

σ(K)U by [step 1.1].

step 1.1
2.2

Claim 2: let λ0 and choose n with 1/nλ/2 by [A2]. Every zB(λ,λ/2) satisfies zλzλ>λ/21/n, so the set F:=σ(K)B(λ,λ/2) is contained in S1/n and is finite by [step 1.1]. Put E:=F{λ}. If E is nonempty, the finite set of positive numbers {zλ:zE} has a minimum ρ>0; if E is empty, put ρ:=λ/2. For r:=min(ρ,λ/2)>0, any zB(λ,r)σ(K) lies in F, while zλ would put z in E and give the contradiction zλρr>zλ. Hence B(λ,r)σ(K){λ}.

step 1.1A2A4
3.1

Claim 1: σ(K) is at most countable, being a subset of the at most countable set U, by [A3]; moreover U is at most countable by [step 1.2] and σ(K)U by [step 2.1].

step 1.2step 2.1A3
4.1

Finally, if λ0 and every punctured ball around λ met σ(K), then by [step 2.2] the punctured ball B(λ,r){λ} meets σ(K) yet contains none of its points, a contradiction; so the only possible accumulation point is 0, and claims 1 and 2 are [step 3.1] and [step 2.2].

step 3.1step 2.2A4

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

69 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