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Prove there are uncountable sets of real numbers that do not contain uncountable closed subsets. (Hint: one can show the existence of a set X in R that does not contain any subset Y having the properties that Y is closed and contains no isolated points.)

This is a question from the book Integration and Modern Analysis. I guess for the hint, one can assume every set in R contains some subset Y that is closed and contains no isolated points, and contradict it with a counterexample such as {1,2}. How can one go from here?

MathNoob
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Sep 07 '21 at 01:36
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    You'll need the set you build to be uncountable, too, so ${1,2}$ isn't relevant. It turns out that there's no easy way to prove this - the existence of an uncountable set of reals with no uncountable closed subset cannot be proved in $\mathsf{ZF}$ (= usual mathematics without the axiom of choice). You'll need to use transfinite recursion here (with choice being used to set the recursion up in the first place). – Noah Schweber Sep 07 '21 at 02:33
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    Does this answer your question? How to construct a Bernstein set and what are their applications? – Conifold Sep 07 '21 at 05:03

1 Answers1

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The proof of this requires the Axiom of Choice. In particular, without the Axiom of Choice it is consistent that every set of real numbers has the perfect set property, which is equivalent to the negation of your exercise.

To construct an uncountable set of real numbers that does not contain an uncountable closed subset, enumerate the set of all uncountable closed subsets (there are $|\Bbb R|=2^{\aleph_0}$ many such sets) as a transfinite sequence $\langle C_\alpha\mid\alpha<2^{\aleph_0}\rangle$. Then, recursively go through these sets and pick two points $r_\alpha,s_\alpha\in C_\alpha$ that are different from every $r_\beta,s_\beta$ for $\beta<\alpha$. This is always possible, since at stage $\alpha$ of the recursion, you picked less than $2^{\aleph_0}$ many points, and $|C_\alpha|=2^{\aleph_0}$ (and in case the latter is not clear, see for example this answer).

Finally, let $R=\{ r_\alpha\mid \alpha<2^{\aleph_0}\}$ and $S=\{ s_\alpha\mid \alpha<2^{\aleph_0}\}$, then either of these two sets satisfies the property that they are uncountable and do not contain any uncountable closed subset.

Sets like this $R$ are called Bernstein sets, by the way.

Vsotvep
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  • We also need the result that any closed uncountable subset $C_{\alpha}$ of reals has cardinal $2^{\aleph_0}$ so that $C_{\alpha}\setminus \cup_{\beta<\alpha}{r_{\beta},s_{\beta}}$ so that $r_{\alpha}$ and $s_{\alpha}$ can exist. – DanielWainfleet Sep 07 '21 at 03:16
  • Indeed, I believe I mentioned this as $|C_\alpha|=2^{\aleph_0}$, at the end of the second paragraph. – Vsotvep Sep 07 '21 at 03:18
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    Yes but I do not think $|C_{\alpha}|=2^{\aleph_0}$ is obvious and a student reading an intro to analysis would likely not know it nor how to prove it. It's not a flaw in your answer though. – DanielWainfleet Sep 07 '21 at 03:26
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    Agreed again. If this isn't clear, then likely there was no prior treatment of uncountable perfect sets being of size continuum either, or of the Cantor-Bendixson theorem, in which case I personally think this a bit of a sadistic exercise... – Vsotvep Sep 07 '21 at 03:37
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    This question has a good exposition of the size of uncountable closed sets: https://math.stackexchange.com/a/740350/176025 – Vsotvep Sep 07 '21 at 03:42
  • A problem (verbatim) from American Monthly Monthly: "A student asserted that any uncountable set of reals contains a closed uncountable subset. Is this true?" My first thought was it was very likely true that a student had asserted this. – DanielWainfleet Sep 07 '21 at 04:22
  • I like your most recent comments....Any non-empty complete metric space with no isolated points has a subspace homeomorphic to the Cantor set and any uncountable closed set in $\Bbb R$ has an uncountable closed subset with no isolated points. – DanielWainfleet Sep 07 '21 at 04:29