this post was submitted on 27 May 2024
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submitted 5 months ago by flamingos@feddit.uk to c/196@lemmy.blahaj.zone
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[โ€“] i_love_FFT@lemmy.ml 1 points 5 months ago (1 children)

Well that's interesting: in order to define unmeasurable sets, you relied on the axiom of choice... I suppose it might be possible to define unmeasurable sets without AC, but maybe not!

Every time I encounter the axiom of choice implying a bunch of crazy stuff, it always loop back to requiring AC. It's like a bunch of evidence against AC!

I find it interesting that the basic description of AC sounds very plausible, but I'm still convinced mathematicians might have made the wrong choice... (See what i did there? ๐Ÿ˜„)

[โ€“] kogasa@programming.dev 2 points 5 months ago

It's required, but nontrivially so. It has been proven that ZF + dependent choice is consistent with the assumption that all sets of reals are Lebesgue measurable.