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There's only a countable number of formulas of ZFC. Therefore, there's only a countable number ways of writing out a formula of ZFC that uniquely specifies a re
by mag487 13y ago
There's only a countable number of formulas of ZFC. Therefore, there's only a countable number ways of writing out a formula of ZFC that uniquely specifies a real number. Therefore, only countably many real numbers exist that have such definitions. And all countable sets of reals have zero Lebesgue measure.
- nawitus 13y agoAdditional question: what's the Kolmogorov complexity of a irrational numberwhich is not described with a ZFC formula? And how is it described?
- mag487 13y agoI haven't really studied algorithmic information theory, but I'd assume that Kolmogorov complexity isn't defined for uncomputable/undefinable numbers. Or maybe it's defined as "infinite," but either way, my guess is that such numbers are simply ignored. (Interestingly, the function which takes a computable number and outputs its Kolmogorov complexity is itself uncomputable!)
- nawitus 13y agoThe article at least implies that the Kolmogorov complexity of uncomputable irrational numbers is larger than computable irrational numbers.