# 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!

• Note that general regularity properties imply that the answer is yes if $E$ contains a Lebesgue-measurable set $A \subset E$ of positive Lebesgue measure : en.wikipedia.org/wiki/Regularity_theorem_for_Lebesgue_measure (In particular, that works if $E = \mathbb R \setminus \mathbb Q$.) But your general question seems interesting. – PseudoNeo May 18 '15 at 8:32
• @Jared, there are a couple of things wrong with your comment. You are right to infer that an uncountable subset of $\mathbb R$ must contain some irrational numbers, but that's not enough to answer the question in the negative. A set of exactly two irrational numbers is closed, for example, so your comment does not amount to a counterexample. Your second comment has more serious problems. No, that's not at all what "uncountable" means-- look it up. And the countability of the rationals in an interval is not, by any means, a consequence of non-continuity. – alexis May 18 '15 at 10:53

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.
• I don't know if I love your answer (because, objectively, it's perfect) or if I hate it (because I understand $\mathbb R$ a bit less now that I know that such beasts exist)... +1 – PseudoNeo May 18 '15 at 8:40
• @PseudoNeo I don't really have a good reference. I first learned of these in a basic set theory book by Just and Weese. It appears that some questions have been asked here: "bernstein set" is:question. – бір-төрт-төрт-үш-жеті-бес May 18 '15 at 9:03