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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The closure of a bounded image contains a ball

Statement

Assume DC. If $T : X \to Y$ is a surjective bounded linear map between Banach spaces, then for some $r > 0$ ,

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

Facts & Assumptions

Given: DC and a surjective bounded linear map $T : X \to Y$ with $X, Y$ Banach.

[F1]

Dependent Choice is the stated countable-selection premise for the Baire category theorem used below (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · direct

1.1givenF1

Surjectivity gives $Y = ⋃_{n \geq 1} \overline{T (B_{X} (0, n))}$ . Baire applied to $Y$ under the stated DC premise gives $n$ , $y_{0}$ , and $ρ > 0$ with $B (y_{0}, ρ)$ inside this closure.

2.1step 1.1algebra

Subtracting two points in this ball shows $B (0, ρ) \subseteq \overline{T (B (0, 2 n))}$ .

3.1step 2.1algebra∎

Scaling by $2 n$ gives $B (0, ρ / (2 n)) \subseteq \overline{T (B (0, 1))}$ , as required.

Depends on

Used by

Dependency tree · two levels

24 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

Buhler--Salamon, Functional Analysis, Lemma 2.9 (standard reference, not scraped)