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I have recently been working through the consequences assuming the Axiom of Determinacy has and came across the fact that this axiom implies that every uncountable set of reals has the perfect set property. I find it to be pretty amazing that AD implies every subset of the reals is either countable or has a perfect subset. I was wondering what consequences this property has for the real numbers? Does it majorly change anything that we know to be true under the usual axioms of ZFC? I know the main result we get from this is CH, but is there any other interesting things that this implies?

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    $\begingroup$ Well, you get CH. $\endgroup$ – Asaf Karagila Mar 21 '16 at 16:44
  • $\begingroup$ I was aware of this fact, and am wondering if there are any other interesting properties coming from this fact. I edited the question so what I am looking for is more clear. $\endgroup$ – Matt Dyer Mar 21 '16 at 16:49
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    $\begingroup$ Choice fails (there is no $\omega_1$ sequence of distinct reals). $\endgroup$ – Andrés E. Caicedo Mar 21 '16 at 16:55
  • $\begingroup$ And to add to what Andres said, either $\omega_1$ is singular, or there is an inaccessible in $L$. $\endgroup$ – Asaf Karagila Mar 21 '16 at 17:00
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    $\begingroup$ @Matt Yes, $\mathsf{AD}$ implies that choice fails. My point is that you do not need such a strong assumption: just working in $\mathsf{ZF}$, the perfect set property for all sets of reals already implies the failure of choice. Similarly, $\mathsf{AD}$ implies that $\omega_1$ is regular and that there are many inaccessible cardinals in $L$, but Asaf is pointing out that, again, in $\mathsf{ZF}$, the perfect set property is enough to show the dichotomy he indicates. For large cardinals, I suggest you consult Kanamori's book or Cantor's attic. $\endgroup$ – Andrés E. Caicedo Mar 21 '16 at 19:56

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