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CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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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:X→X 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 λ∈C with λ≠0 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:={λ∈σ(K):∣λ∣≥ε} is finite (Riesz schauder spectrum of a compact operator).

[A2]

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

[A4]

In C the balls are those of the metric d(z,w)=∣z−w∣ (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:X→X, and the sets S1/n={λ∈σ(K):∣λ∣≥1/n}, n≥1.

1.1A1A2

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

1.2A1A3

The set U:={0}∪⋃n≥1S1/n is at most countable: it is a countable union of finite sets, and [A3] applies.

2.1step 1.1

σ(K)⊆U by [step 1.1].

2.2step 1.1A2A4

Claim 2: let λ≠0 and choose n with 1/n≤∣λ∣/2 by [A2]. Every z∈B(λ,∣λ∣/2) satisfies ∣z∣≥∣λ∣−∣z−λ∣>∣λ∣/2≥1/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 {∣z−λ∣:z∈E} has a minimum ρ>0; if E is empty, put ρ:=∣λ∣/2. For r:=min⁡(ρ,∣λ∣/2)>0, any z∈B(λ,r)∩σ(K) lies in F, while z≠λ would put z in E and give the contradiction ∣z−λ∣≥ρ≥r>∣z−λ∣. Hence B(λ,r)∩σ(K)⊆{λ}.

3.1step 1.2step 2.1A3

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].

4.1step 3.1step 2.2A4∎

Finally, if λ≠0 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].

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