Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07
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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:(c00,1)(c00,).

Counterexample

technique · direct
1.1

Since xx1, I is bounded and bijective.

given
1.2

Its inverse is unbounded: for x(m)=(1/m,,1/m,0,), x(m)=1/m but x(m)1=1.

given
1.3

The target is incomplete by Uniform boundedness fails on the incomplete space c_00. The partial sums of (2k)k1 are also Cauchy in the 1 norm but have no limit in c00, so the domain is incomplete as well.

given
2.1

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

step 1.2

Depends on

Used by

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