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by dwohnitmok
367 days ago
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This is subtle and a simple counting argument (definable means satisfies a finite formula, there are only countably many finite formulas, there are uncountably many reals, therefore there must be undefinable reals) doesn't work, because "definable in ZFC" is not something that is formalizable in ZFC and so the usual set-theoretic counting arguments don't work. So it is in fact possible and consistent with ZFC that all reals are definable. See: https://mathoverflow.net/questions/44102/is-the-analysis-as-... |
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