Prove or disprove :
The set of limit points of an uncountable set of real numbers is uncountable .
This is easier said in more general terms.
Lemma: If is second countable and is covered by then omits a countable subcover .
Proof:Call the countable open base . For each there exists some set such that . But, we have that there exists some such that . Clearly then the collection covers and is countable. Taking an which contains for each finishes the argument.
Now we can prove the problem at hand.
Let be second countable and uncountable, then is uncountable.
Assume not. Then, is countable and so is uncountable. But, since each point is in particular not a limit point of there exists a neighborhood such that . Clearly the class is an open cover for and so by the lemma it must admit a finite subcover . But, but since each is a singleton it follows that is countable Contradiction.
Your problem is a corollary since is a dense subset of and every separable metric space is second countable.
Figured it out just too late
Thanks for the proof, I'll give it the attention which it deserves when I'm not dead tired! Exams begin tomorrow.
(After all, what you had written before worked also. Why did you remove it? Every point of is an isolated point, and so has the cardinality of a collection of disjoint open sets, which is countable.)
Because the needn't be disjoint. I was being stupid. Clearly they can intersect but they cannot contain each other. For example could be two of the sets if [math[\frac{1}{2},2\in E[/tex] since they're intersection with could easily be a singleton but they aren't disjoint.