Given any uncountable subset $S$ of $\mathbb{R}$ , does there exist a non empty perfect set $A$ such that $A \subseteq S$ ?
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What's a perfect set? – Henrik supports the community Jun 03 '17 at 08:48
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I think that $S =\mathbb{R}\setminus \mathbb{Q}$ does not contain any nonempty perfect set. – Rigel Jun 03 '17 at 08:50
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It has...problem in Rudin ...chapter 2 qstn 18 – CoffeeCCD Jun 03 '17 at 08:52
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https://math.stackexchange.com/questions/1064/perfect-set-without-rationals?rq=1 – CoffeeCCD Jun 03 '17 at 09:13
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Oh yes, sorry, you are right! – Rigel Jun 03 '17 at 09:15
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In these terms, I think it can only be proved false: it would imply that any uncountable (id est, with cardinality $\ge\aleph_1$) subset of $\Bbb R$ has cardinality $2^{\aleph_0}$, which is basically CH. I don't know for the same question with "cardinality of the continuum" instead of "uncountable". – Jun 03 '17 at 09:21
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It depends on the use of the axiom of choice AC.
a subset of a Polish space has the perfect set property if it is either countable or has a nonempty perfect subset (Kechris 1995, p. 150).
The axiom of choice implies the existence of sets of reals that do not have the perfect set property.
This says that in a polish space (like the real numbers) AC implies, there are sets without the perfect set property, which means they are not countable and do not contain a perfect subset.
However, in Solovay's model, which satisfies all axioms of ZF but not the axiom of choice, every set of reals has the perfect set property, so the use of the axiom of choice is necessary.
