CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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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.
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A bounded bijection of incomplete normed spaces need not be open

Statement refuted

A bounded bijective linear map between arbitrary normed spaces need not be open.

Facts & Assumptions

Given: The identity $I : (c_{00}, ∥ \cdot ∥_{1}) \to (c_{00}, ∥ \cdot ∥_{\infty})$ .

Counterexample

technique · direct

1.1given

Since $∥ x ∥_{\infty} \leq ∥ x ∥_{1}$ , $I$ is bounded and bijective.

1.2given

Its inverse is unbounded: for $x^{(m)} = (1 / m, \dots, 1 / m, 0, \dots)$ , $∥ x^{(m)} ∥_{\infty} = 1 / m$ but $∥ x^{(m)} ∥_{1} = 1$ .

1.3given

The target is incomplete by Uniform boundedness fails on the incomplete space c_00. The partial sums of $(2^{- k})_{k \geq 1}$ are also Cauchy in the $ℓ^{1}$ norm but have no limit in $c_{00}$ , so the domain is incomplete as well.

2.1step 1.2∎

If $I$ were open, its inverse would be continuous at $0$ , hence bounded by linearity, contradicting step 1.2.

Depends on

Used by

A closed everywhere-defined graph need not be bounded without completeness Counterexample

Dependency tree · two levels

6 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, Example 2.13 (standard reference, not scraped)