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 $x \in H$ .

1.1A1algebra

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.

2.1step 1.1A2A3

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 $A_{m}$ .

3.1step 2.1A4

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.

4.1step 3.1A5A6∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources