# Another second countable space problem

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## Homework Statement

I'm pretty sure I got this one right, but I'd still like to check, as always.

Let X have a countable basis, and let A be an uncountable subset of X. Show that uncountably many points of A are limit points of A.

## The Attempt at a Solution

Now, since X is second-countable, A must be second-countable, too. Hence, there exists a countable subset B of A which is dense in A, i.e. Cl(B) = A. Now, Cl(B) = B U B', where B' is the set of all limit points of B. Since B is countable it follows that B' must be uncountable, and B' is exactly the set of all limit points of A, too.

Yes, this is correct. But it's a bit fishy since you used two different closures in there, the first time you took the closure in B and the second time you took the closure in X. But the argument is essentially correct.

Also note that we really needed second countabilty. Separability of the space would not have been enough, since separability only transfers to open subspaces...

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Hm, I don't really see how the "closure in B" affects the whole thing? You mean, since B is a subspace of A, Cl(B) is regarded as a closure in A?

Yes, I see that separability is weaker in general.

No, it doesnt affect the only thing. I was only a bit confused when I read your proof, but it works fine!
Forget what I said

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OK, thanks (as always)!