LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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.

Successive approximation turns a closure-ball inclusion into an actual preimage

Statement

Assume DC. Let $T : X \to Y$ be bounded linear, with $X$ Banach. If $B_{Y} (0, r) \subseteq \overline{T (B_{X} (0, 1))}$ for some $r > 0$ , then

$B_{Y} (0, r / 2) \subseteq T (B_{X} (0, 1)) .$

Facts & Assumptions

Given: DC and $T, X, Y, r$ with the displayed closure inclusion.

[F1]

Dependent Choice licenses a sequence of witnesses when each next choice depends on the current residual (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · constructive

1.1givenF1construct

Given $y$ with $∥ y ∥ < r / 2$ , choose a real $θ$ with $∥ y ∥ / r < θ < 1 / 2$ . Repeatedly use the closure inclusion after scaling to choose $x_{n}$ with $∥ x_{n} ∥ < θ 2^{- n}$ and residual $y - T \sum_{k \leq n} x_{k}$ of norm $< r θ 2^{- n - 1}$ . The initial residual has norm below $r θ$ ; each later residual lies in the ball to which the next scaled closure inclusion applies. [F1] licenses this dependent recursive selection.

2.1step 1.1

The series $\sum_{n} x_{n}$ converges in $X$ by Series criterion for Banach spaces, and its sum $x$ has $∥ x ∥ \leq \sum_{n} ∥ x_{n} ∥ \leq 2 θ < 1$ .

3.1step 1.1step 2.1discharge-construct∎

Boundedness makes $T$ continuous by For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, so $T x = \lim_{n} T \sum_{k \leq n} x_{k} = y$ as the residuals vanish.

Depends on

Dependency tree · 2 levels

Sources

Buhler--Salamon, Functional Analysis, Lemma 2.10