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

1. Dec 26, 2011

### nonequilibrium

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?

2. Dec 26, 2011

### jgens

3. Dec 26, 2011

### jgens

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.

Last edited: Dec 26, 2011
4. Dec 26, 2011

### nonequilibrium

What a pity!

Thank you :)

5. Dec 27, 2011

### Bacle2

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.

Last edited: Dec 27, 2011