Post #2566014
2026-04-19 15:06 UTC
@Arpie4Math@mathstodon.xyz Nice write-up. So, here's the question:
Axiom of Choice breaks measure theory. One then concludes that asserting every set has a well-defined measure (excluding 0 & ∞) implies...what? Does it mean that measure theory requires that mysterious border material? Does it mean that measure theory needs non-orderable sets?
Is this an open problem? Measure theory + AC breaks when there is a group symmetry, so simple symbolic logic may not be sufficient. Or is that the clue - a group symmetry makes something non-orderable?
Replies (1)
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@Arpie4Math@mathstodon.xyz 2026-04-19 16:58
@skewray@mathstodon.xyz Does AC break measure theory, or is measure theory already broken with the idea that all sets have a measure? That is, the axiom of choice allows one to reason about sets (Vitali sets) that are paradoxical with respect to the assumptions of measure theory but negating the axiom of choice does nothing to justify the assumption that all sets of real numbers have a measure. https://enrichedjamsham.substack.com/p/the-axiom-of-choice-is-not-controversial