Every uncountable subset of R has a limit point in Q?

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Discussion Overview

The discussion centers around the question of whether every uncountable subset of the real numbers \( \mathbb{R} \) has at least one limit point in the rational numbers \( \mathbb{Q} \). Participants explore this concept through various mathematical arguments and counterexamples, engaging in both theoretical reasoning and specific constructions.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that every uncountable subset \( A \subset \mathbb{R} \) should have a limit point in \( \mathbb{Q} \) but expresses uncertainty about how to prove this.
  • Another participant asserts that the claim is false and references a counterexample involving a non-empty perfect set that contains no rational points.
  • A third participant recalls a method involving a Cantor-set type construction, suggesting that by removing intervals containing rational numbers, one could create a non-empty perfect set that is uncountable.
  • This participant also mentions the possibility of detailing the construction further if needed, indicating a willingness to clarify their approach.
  • Another contribution states that every uncountable subset of \( \mathbb{R}^n \) does have a limit point, referencing the Lindelöf property of \( \mathbb{R}^n \) and suggesting that one could potentially adjust the set to ensure the limit point is irrational.

Areas of Agreement / Disagreement

Participants express disagreement regarding the original claim, with some asserting it is false and providing counterexamples, while others maintain that there should be a limit point in \( \mathbb{Q} \). The discussion remains unresolved with multiple competing views presented.

Contextual Notes

Participants reference various mathematical properties and constructions, but the discussion includes unresolved assumptions and lacks a definitive resolution regarding the existence of limit points in \( \mathbb{Q} \) for uncountable subsets of \( \mathbb{R} \).

nonequilibrium
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Hello,

I would expect that every uncountable [itex]A \subset \mathbb R[/itex] has (at least) one [itex]q \in \mathbb Q[/itex] as a limit point of A. I don't really know how to prove this, though. I have the feeling that it shouldn't be too hard. Can someone get me on the right track?
 
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If I recall correctly, I think the method that I worked out involved a cantor-set type construction: At the nth step you remove small open intervals containing all rational numbers of the form p/n in [0,1]. By choosing the length of the intervals properly, you can ensure that the end points of these intervals remain in the set.

If you do this correctly, I think you should be able to show that the set is non-empty and perfect, which implies that it is uncountable. I would have to work out all of the details again to make sure I am remembering this correctly, but I think this is another way of doing it.

Edit: I think one way of choosing the length of the intervals is to ensure that at the nth step of this construction, you remove at most as much as you do in the nth step of the construction of the cantor set. I realize this is kind of an incomprehensible mess, so if you want more details, I should be able to write out the construction for you.
 
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What a pity!

Thank you :)
 
I know something related is true:

Every uncountable subset of ℝn does have a limit point.

This comes from the fact that ℝn is Lindeloff (every cover has a countable
subcover): cover your Euclidean space by the (countable-this is key) concentric balls B(0,n)
Then, by cardinality arguments, one of your bounded balls has infinitely-many elements of the set . Now, use Weirstrass result that every bounded infinite subset of ℝn has a limit point.

Maybe you can choose/tweak the set so that your limit point is irrational.
 
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