From a wellordering of an uncountable set of reals, Bernstein constructed a set of reals without the perfect set property. My question is, does an uncountable well-ordering itself violate the perfect set property? Equivalently, if $W \subset \mathbb{R}^2$ is a well-ordering of some set of reals, must $W$ be thin?
2026-03-27 02:34:28.1774578868
Well-orderings and the perfect set property
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Since you said, in a comment, "you can assume choice if necessary", I'll assume choice and fix a well-ordering $W$ of all the reals such that all the negative reals precede all the positive reals. Then $W$ includes the perfect set $\{(x,y):x<0<y\}$.
On the other hand, some such use of the axiom of choice is needed. It's consistent with ZF + DC that every well-orderable set of reals is countable.