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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!

JiK
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user119882
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    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 : http://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 08:32
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    @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

1 Answers1

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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:

Community
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Taumatawhakatangihangakoauauot
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    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 08:40
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    I wouldn't like to bother you (that would be a very poor reward for your answer) but the Wikipedia article really frustrates me. What is your favourite reference for the existence of Bernstein sets? – PseudoNeo May 18 '15 at 08:56
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    @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. – Taumatawhakatangihangakoauauot May 18 '15 at 09:03
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    Good answer. It might be worth adding that every uncountable Borel set contains an uncountable closed set. (Same goes for an uncountable analytic set, whatever that is.) So, any well-defined set that you run into in everyday life, if it's uncountable, it contains an uncountable closed set. The counterexamples are monstrosities that are "constructed" with the aid of the axiom of choice. – bof May 18 '15 at 09:48
  • @bof: Thank-you for your suggestion. – Taumatawhakatangihangakoauauot May 18 '15 at 10:12
  • I am not familiar with these named sets, I wonder if those sets, for example, if there exists one Borel set which contains no rational number and uncountable in theoretically or by given a concrete example, just for curious, thanks! – user119882 May 18 '15 at 13:16
  • @user119882 See Wikipedia for a basic introduction to Borel sets. This collection contains all open sets, and is closed under complementation and countable unions/intersections. In particular the set of all irrationals is Borel. Almost every subset of $\mathbb{R}$ you come across will be Borel. – Taumatawhakatangihangakoauauot May 18 '15 at 15:30