Are the perfect-free sets countable?

Question

Let $A$ be subset of $\mathbb{R}$ that contains no nonempty perfect subsets. Is $A$ countable?

Answer 1

No. If $\aleph_1<2^{\aleph_0}$ then uncountable set of cardinality smaller than $2^{\aleph_0}$ does not contain a perfect subset. But more generally we can construct something called a Bernstein set which has size $2^{\aleph_0}$ but does not contain a perfect subset.

The construction is by transfinite induction and it uses a well-ordering of $2^{\aleph_0}$. And indeed it is consistent that the axiom of choice fails, and every uncountable set of real numbers contains a perfect subset.

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Related questions:

• A set which is neither meagre nor comeagre in any interval.

Answer 2

On the other hand, assuming the Axiom of Determinacy (which contradicts the axiom of choice) every subset of a Polish space has the Perfect Set Property, namely it is either countable or it contains a Cantor set.