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

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Every uncountable subset of the real numbers does not necessarily have a limit point in the rationals, as demonstrated by a counter-example involving a Cantor-set type construction that avoids rational points. The construction involves removing small open intervals containing rational numbers while ensuring the remaining set is non-empty and perfect, thus uncountable. The discussion also notes that while uncountable subsets of ℝn do have limit points, this property does not extend to rational numbers specifically. The method relies on the Lindelöf property of ℝn and the Weierstrass theorem, which states that every bounded infinite subset has a limit point. Overall, the initial assumption about uncountable subsets having rational limit points is proven false.
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Hello,

I would expect that every uncountable A \subset \mathbb R has (at least) one q \in \mathbb Q 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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