False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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False: a limit-computable function has a known stabilization stage

Statement

Every limit-computable function has a computable modulus of stabilization.

Facts & Assumptions

Given: the standard stage approximation to the halting set.

Refutation

technique · contradiction

1.1givenconstruct

Let $h (e, s) = 1$ if program $e$ has halted by stage $s$ , and $0$ otherwise. Bounded simulation makes $h$ total computable, and it converges pointwise to the characteristic function $χ_{0^{'}}$ of the halting set. Thus $χ_{0^{'}}$ is limit computable.

2.1step 1.1assume-contradischarge-contradiction∎

Suppose $χ_{0^{'}}$ had a total computable approximation $g (e, s)$ with a computable stabilization modulus $m (e)$ . Computing $g (e, m (e))$ would then decide $0^{'}$ , contradicting undecidability of the halting problem. This argument covers every possible approximation, including $h$ from step 1.1, so no such modulus exists.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Ludovic Patey, Computability Theory, §4.7 (standard reference, not scraped)