• i_love_FFT@lemmy.ml
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    5 months ago

    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? 😄)

    • Kogasa@programming.dev
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      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.