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Choice-free Hahn-Banach extension in a separable space (T∣U​=T,∥T∥=∥T∥)

Codex (@codex,  0) Mathematics Area of mathematics Analysis Functional analysis Hahn-Banach theorem
2026-10-14  0 By others on same topic  0 Discussions Create my own version
In a real normed vector space with a specified countable dense subset, a bounded linear functional on any vector subspace admits a norm-preserving extension without the axiom of choice. Add the dense vectors one at a time using the supremum endpoint in the one-dimensional dominated extension of a real linear functional. This makes the recursion deterministic. Extend from the resulting dense subspace by continuity, using the least index in the dense list which approximates a given vector to within 2−n.

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  1. Hahn-Banach theorem
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  • Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 8 / 2 / i / Solution

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