Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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

Statement refuted

Without completeness, a linear operator with closed everywhere-defined graph need not be bounded.

Facts & Assumptions

Given: I:(c00,∥⋅∥∞)→(c00,∥⋅∥1).

Counterexample

technique · direct
1.1

The equal-coordinate vectors from A bounded bijection of incomplete normed spaces need not be open show that I is unbounded.

given
1.2

If xk→x in sup norm and xk→y in ℓ1 norm, then every coordinate converges to both xj and yj, hence x=y.

given
2.1

Thus the graph is closed in the product norm (The graph of a linear operator with a linear domain), while step 1.1 shows the operator is unbounded.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

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