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A special case of a theorem of Brian Scott (from On the existence of totally inhomogeneous spaces) is that there is a size-continuum set $S\subset\mathbb{R}$ such that if $x,y\in S$ are distinct then $S\setminus\{x\}\not\cong S\setminus\{y\}$. The proof relies heavily on the axiom of choice, and at a glance any such $S$ must be quite wild. However, I don't immediately see how to show that such an $S$ must fail any of the standard regularity properties (e.g. measurability).

Question: Is the existence of such an $S$ consistent with $\mathsf{ZF+DC+AD}$?

"Obviously" the answer is negative ...

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    $\begingroup$ A related question (I do not know the answer either): Is there an infinite Borel subset of $ℝ$ that does not have a nontrivial auto-homeomorphism? $\endgroup$
    – Dmytro Taranovsky
    Commented Jul 7, 2023 at 22:05
  • $\begingroup$ It appears that the question was answered by van Engelen, Miller, and Steel, see this review. $\endgroup$
    – KP Hart
    Commented Jul 8, 2023 at 12:52
  • $\begingroup$ @KPHart Delightful! If you add that as an answer, I'll accept it. $\endgroup$
    – Noah Schweber
    Commented Jul 8, 2023 at 13:12
  • $\begingroup$ What structure does the $\cong$ refer to? $\endgroup$
    – Emil Jeřábek
    Commented Jul 8, 2023 at 13:48
  • $\begingroup$ @EmilJeřábek Homeomorphism as topological spaces. (See Scott's paper.) $\endgroup$
    – Noah Schweber
    Commented Jul 8, 2023 at 13:51

1 Answer 1

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In Rigid Borel sets and better quasiorder theory (Logic and combinatorics, Proc. AMS-IMS-SIAM Conf., Arcata/Calif. 1985, Contemp. Math. 65, 199-222 (1987), zbMath review here) Fons van Engelen, Arnold Miller, and John Steel showed that the only rigid Borel sets are the singletons; the proof yields the same for all sets of reals if one assumes $\mathsf{AD}$. So the answer to the present question is negative.

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