LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Only countably many coefficients of a square-summable family are nonzero

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

If $a = (a_{i})_{i \in I} \in ℓ^{2} (I, F)$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ), then its support ${i \in I : a_{i} \neq 0}$ is at most countable (Finite, countably infinite, countable, uncountable).
If $(e_{i})_{i \in I}$ is an orthonormal family in a real or complex inner-product space $H$ and $x \in H$ , then the set ${i \in I : ⟨ x, e_{i} ⟩ \neq 0}$ of nonzero coefficients of $x$ is at most countable.

The hypothesis is not decoration. The countable-union step below selects one surjection of $N$ onto each of the countable sets in a countable family, which is exactly the Axiom of Countable Choice; the threshold sets themselves and their finiteness are ZF.

Facts & Assumptions

[A1]

$ℓ^{2} (I, F)$ consists of the families with finite square sum $S = \sum_{i \in I} ∣ a_{i} ∣^{2}$ , the sum being the supremum of the finite subsums; if $S < + \infty$ then for every real $ε > 0$ there is a finite tail-control set $F$ with $\sum_{i \in I ∖ F} ∣ a_{i} ∣^{2} < ε$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[A2]

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

[A3]

A subset of an at most countable set is at most countable, and finite sets are at most countable (Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[A4]

Under the Axiom of Countable Choice, a countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[A5]

For $x \in H$ and an orthonormal family $(e_{i})_{i \in I}$ , the square sum $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ is finite (The Bessel inequality for an arbitrary orthonormal family).

[A6]

A family belongs to $ℓ^{2}$ exactly when its square sum is finite (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

Proof

technique · direct

Given: Countable Choice, and first a family $a \in ℓ^{2} (I, F)$ with $S = \sum_{i \in I} ∣ a_{i} ∣^{2} < + \infty$ ; for the second claim an orthonormal family $(e_{i})_{i \in I}$ in $H$ and $x \in H$ .

1.1

For each natural $m \geq 1$ put $A_{m} : = {i \in I : ∣ a_{i} ∣ \geq 1 / m}$ . Since $S$ is finite, choose a finite tail-control set $F_{m} \subseteq I$ with $\sum_{i \in I ∖ F_{m}} ∣ a_{i} ∣^{2} < 1 / m^{2}$ ; then $A_{m} \subseteq F_{m}$ , because $i \notin F_{m}$ would give $1 / m^{2} \leq ∣ a_{i} ∣^{2} \leq \sum_{j \in I ∖ F_{m}} ∣ a_{j} ∣^{2} < 1 / m^{2}$ , a contradiction.

A1algebra

2.1

Each $A_{m}$ is a subset of the finite set $F_{m}$ , hence is at most countable, and the support of $a$ satisfies ${i : a_{i} \neq 0} = ⋃_{m \geq 1} A_{m}$ : if $a_{i} \neq 0$ then $∣ a_{i} ∣ > 0$ and [A2] provides $m \geq 1$ with $1 / m < ∣ a_{i} ∣$ , that is $i \in A_{m}$ ; the reverse inclusion is immediate from the definition of $A_{m}$ .

step 1.1A2A3

3.1

The union ${i : a_{i} \neq 0} = ⋃_{m \geq 1} A_{m}$ is a countable union of at most countable sets, indexed by the natural numbers $m \geq 1$ , so it is at most countable by [A4]; this proves the first claim.

step 2.1A4

4.1

For the second claim, apply the Bessel inequality to the coefficient family of $x$ : its square sum is finite, so that family lies in $ℓ^{2} (I, F)$ by [A6], and the first claim now shows that the set of indices with $⟨ x, e_{i} ⟩ \neq 0$ is at most countable.

step 3.1A5A6∎

Depends on

Used by

Fourier expansion in a Hilbert space Theorem

Dependency tree · two levels

51 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

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §2.1, p.48, threshold argument after (2.4) (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, printed pp.72–80 (standard reference, not scraped)