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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 be a complex Banach space, let be a compact operator (Compact linear operator) and let be its spectrum (Spectrum and resolvent of a bounded operator). Then:
- is at most countable (Finite, countably infinite, countable, uncountable);
- for every with there is a real with (Open ball, closed ball and sphere in a metric space).
In particular the only point of that can be an accumulation point of (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) is .
Facts & Assumptions
Under AC, for every real the set is finite (Riesz schauder spectrum of a compact operator).
For every real there is a natural with (For every in a complete ordered field there is a natural with ).
Under , an at most countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming , The Axiom of Countable Choice (), Finite, countably infinite, countable, uncountable), and every subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable); implies (AC supplies the countable and dependent choices used in Banach integration, The Axiom of Choice).
In the balls are those of the metric (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane, Open ball, closed ball and sphere in a metric space); a point is an accumulation point of a set when every punctured ball meets (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).
Proof
Given: , a complex Banach space , a compact , and the sets , .
Every subspace of of the form is finite by [A1]; and every with lies in some , because and [A2] gives with .
The set is at most countable: it is a countable union of finite sets, and [A3] applies.
by [step 1.1].
Claim 2: let and choose with by [A2]. Every satisfies , so the set is contained in and is finite by [step 1.1]. Put . If is nonempty, the finite set of positive numbers has a minimum ; if is empty, put . For , any lies in , while would put in and give the contradiction . Hence .
Claim 1: is at most countable, being a subset of the at most countable set , by [A3]; moreover is at most countable by [step 1.2] and by [step 2.1].
Finally, if and every punctured ball around met , then by [step 2.2] the punctured ball meets yet contains none of its points, a contradiction; so the only possible accumulation point is , and claims 1 and 2 are [step 3.1] and [step 2.2].
Depends on
- Riesz schauder spectrum of a compact operator
- Spectrum and resolvent of a bounded operator
- Compact linear operator
- Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$
- Every subset of an at most countable set is at most countable
- For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$
- Finite, countably infinite, countable, uncountable
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- The Axiom of Choice
- AC supplies the countable and dependent choices used in Banach integration
- Open ball, closed ball and sphere in a metric space
- Interior, closure, boundary, limit point, isolated point and dense subset of a metric space
- The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane
- Banach space
Used by
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Sources
- Theo Bühler and Dietmar Salamon, Functional Analysis — §5.2.3 p.225, Theorem 5.21 and the countability remark (standard reference, not scraped)
- Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §6.6, spectrum of a compact operator (standard reference, not scraped)