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So I've been thinking about the axiom of choice (AC) and the axiom of determinacy (AD), which are known to be inconsistent with each other. Some qualitative similarities of them are:

  • they are both trivial in the 'finite' case
  • they have useful consequences (the AC arguably more so, but e.g. AD implies that all subsets of $\mathbb R$ are measurable, so for the sake of the argument lets say they are both useful)

Now I wonder: the axiom of choice has also seemingly paradoxical (e.g. Banach-Tarski) or seemingly 'too strong' implications (e.g. the well ordering theorem). Is there a similar example concerning the axiom of determinacy? I.e. is there an example of a consequence of AD which in some sense feels paradoxical or too strong?

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    $\begingroup$ The fact that all subsets of $\Bbb R$ are measurable seems too strong if not paradoxical. $\endgroup$ – Arnaud Mortier Nov 6 '19 at 17:51
  • $\begingroup$ In my view this feels more like Tychonoffs theorem, i.e. that all products of compact spaces are compact, which seems too strong, but is just so useful. But of course that is a matter of opinion. $\endgroup$ – Redundant Aunt Nov 6 '19 at 18:18
  • $\begingroup$ Well, even if you restrict Tychonoff to Hausdorff spaces, you still contradict AD. $\endgroup$ – Asaf Karagila Nov 6 '19 at 18:20
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Well. You can partition the real numbers into more non empty sets than points. In fact, there is a very long sequence of partitions which refine each other, starting from the reals, and each has more cells than the previous one.

If that's not counterintuitive, I don't know what is.

Under additional assumptions you also have that every countable partial order can be embedded into the cardinals between the reals and their power set. Again if this is not odd, you need to define better what would these weird situations be.

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  • $\begingroup$ Thanks for the answer! Could you elaborate on your first point, i.e. this partition into more sets than points? Where can I read about it? $\endgroup$ – Redundant Aunt Nov 6 '19 at 18:16
  • $\begingroup$ This comes from the fact that AD implies the reals can be mapped onto many ordinals, but the only injections from an ordinal into the reals are from countable ordinals. $\endgroup$ – Asaf Karagila Nov 6 '19 at 18:19

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