Does every uncountable subset of $\mathbb{R}$ have an uncountable closed subset?

Let $E\subseteq \mathbb{R}^1$ be an uncountable set. Can we obtain some subset $F\subseteq E$ which is closed and uncountable?

Basically, I want to construct some set containing only irrational numbers which is also uncountable and closed, in a sense, I want to know a general process to construct such sets.

Thank you!

Solution 1:

In general this cannot be done. This is because of the existence of Bernstein sets. $B \subseteq \mathbb{R}$ is a Bernstein set if both $B$ and $\mathbb R \setminus B$ have nonempty intersection with every uncountable closed set. In particular such sets do not contain any uncountable closed subset.

Of course, such sets are not very pleasant: they do not have the Baire property, and they are not Lebesgue measurable. For Borel subsets of $\mathbb R$ we have a positive result in the form of the perfect set property: every Borel subset of $\mathbb R$ is either countable, or contains a perfect subset.

For the specific case of the irrational numbers, there are several constructions given in this previous question:

• Uncountable closed set of irrational numbers