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This question came from (London Mathematical Society Student Texts) Krzysztof Ciesielski-Set Theory for the Working Mathematician-Cambridge University Press. Chapter 6.2 Exercise 5.

I have thought about it for a few weeks and asked some friends.

The extension question is whether every Borel subset of $\mathbb R$ is either countable, or contains a perfect subset?

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    $\begingroup$ Why did you remove the "descriptive set theory" and "Borel sets" tags? They're obviously appropriate. $\endgroup$ – Noah Schweber May 21 '19 at 12:01
  • $\begingroup$ I wanna a proof just use the method of Set theory , never use analytic set or the method of Descriptive set theory. I don't know whether it could be proved like this or not. Thanks. $\endgroup$ – kan3 May 21 '19 at 22:46
  • $\begingroup$ First, that doesn't justify removing the "borel-sets" tag, which you also did. Moreover, remember that tags are advertisements: even if you want a descriptive-set-theory free answer (and I'm not sure what such a thing would even be, and insofar as I am I strongly suspect none exists), the descriptive-set-theory tag is still relevant since the question itself is a descriptive-set-theoretic one and descriptive set theorists are part of the audience you want to attract. $\endgroup$ – Noah Schweber May 22 '19 at 7:03
  • $\begingroup$ Thank you! Good advice! It is my first time to ask a question.Learn more from you. $\endgroup$ – kan3 May 22 '19 at 11:15
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Yes - this is called the perfect set property. The analytic (= continuous image of Borel) sets also have this property; for more complicated sets (e.g. complements of analytic sets), the perfect set question is undecidable from the usual axioms of set theory.

While it's trivially true for open sets, already for closed sets it takes some work (the easiest approach, given a closed set $C$, is to consider the set of elements of $C$ around which $C$ is "locally uncountable" - that is, those $c\in C$ such that every open $U$ containing $c$ has uncountable intersection with $C$).

The full result follows from Borel determinacy, which is a very hard theorem; off the top of my head I don't know a proof that doesn't use this. Kechris' book Descriptive set theory is a very good source (as is Moschovakis' book).

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    $\begingroup$ @kan3 This is why the continuum hypothesis seemed so plausible to Cantor and contemporaries: every set you can "define" (by logical formulas etc.) in the reals somehow is either size continuum or countable. $\endgroup$ – Henno Brandsma May 21 '19 at 6:03
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    $\begingroup$ There is a variant of the Cantor-Bendixson argument which even shows the PSP for analytical sets which is much easier than Borel determinacy. It also gives more information as it shows generally $\kappa$-Suslin sets have the $\kappa$-PSP. This proof can be found in the second chapter of Moschovakis book. $\endgroup$ – Andreas Lietz May 21 '19 at 7:25
  • $\begingroup$ The other relevant idea (also used in Borel determinacy) is to change the topology by turning the relevant Borel set into a clopen set. Then you apply Cantor-Bendixson. This is in Kechris. $\endgroup$ – Pedro Sánchez Terraf May 21 '19 at 13:30
  • $\begingroup$ I wanna a proof just use the method in Set theory , never use analytic set or the method of Descriptive set theory. I don't know whether it could be proved like this or not. Thanks. $\endgroup$ – kan3 May 21 '19 at 22:49
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    $\begingroup$ You are certainly right, there are such examples - arguably the fundamental theorem of algebra does not have a known purely algebraic proof. But I think in general, this is not to be expected, after all the methods in a field have been developed to tackle the problems encountered there. $\endgroup$ – Andreas Lietz May 22 '19 at 11:50
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I have an idea mainly from Noah Schweber's advice. I don't know is it exactly right or not?

The simple idea is a subset of a Polish space (like the real numbers) has the perfect set property if it is either countable or has a nonempty perfect subset (Kechris 1995, p. 150).

A proof from Noah Schweber' idea is as follow:

Firstly, it is trivially true for open sets, (since the definition of borel set is from the open set). If a borel subset of R is/or contains an open set, we may let it be $B(p, \epsilon)$,then in $R$ there is a $\delta < \epsilon$, such that $[p-\delta, p+\delta]$ is a close set ,and this closed set has no isolated point,and this set is perfect set. From the definition of perfect set, that is, in the field of topology, a subset of a topological space is perfect if it is closed and has no isolated points.

But for closed sets it takes some difficult work. The easiest approach, given a closed set $C$, if it doesn't contain a perfect set ,that is every point/element of $C$,let it be $c \in C$ is isolated points, but this is impossible. Because every family U of pairwise-disjoint open subsets of R is at most countable. For every isolated point of $C$, we could find an enough "small" open set $B(c, \epsilon)$, to cover the isolated point, and we could ensure all the open sets are pairwise-disjoint. So we have only at most countable isolated points. But this closed set of Borel subset is uncountable. So this closed set contains a part of uncountable closed set.

Then from (Cantor–Bendixson theorem) ,every uncountable, closed subset F of R can be represented as a disjoint union of a perfect set P and an at most countable set C.So we have a perfect set.

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    $\begingroup$ And what about for sets that are neither open nor closed? $\endgroup$ – Alex Kruckman May 24 '19 at 0:59
  • $\begingroup$ The problem is when you try to connect closed sets and Borel sets ("but this closed set of Borel subset is uncountable"). There's simply no bridge here. Why do you think that an uncountable Borel set contains an uncountable closed set? You might try to get around this by looking at the closure of a given Borel set, but now there's no way to relate the cardinalities: e.g. $\mathbb{Q}$ is Borel and countable but its closure has size continuum. So as I said in my answer, merely proving the result for closed sets doesn't help us at all (although it is an interesting argument in its own right). $\endgroup$ – Noah Schweber May 24 '19 at 1:26
  • $\begingroup$ And remember that there are Borel sets which are extremely far from open or closed: the Borel hierarchy goes all the way up to $\omega_1$, and open/closed sets live at the very first level. $\endgroup$ – Noah Schweber May 24 '19 at 1:27
  • $\begingroup$ Ok! Thanks a lot for your explanation. I have read Kechris' book in which the Theorem 13.6 gives a proof of the perfect set theorem for Borel Sets by proving an uncountable subset of Polish contains a homeomorphic copy of Cantor space. Does these mean that every uncountable Borel subset of R contain a Cantor subset? Is it a new question? $\endgroup$ – kan3 May 24 '19 at 2:21
  • $\begingroup$ @kan3 "Does these mean that every uncountable Borel subset of R contain a Cantor subset?" Yes. (And indeed the way you prove that all perfect sets have size continuum is by showing that the Cantor set embeds into any perfect subset.) $\endgroup$ – Noah Schweber May 24 '19 at 14:14

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