Let be a non-empty countable set. Suppose has no isolated points, prove that is dense in
It is difficult, isn't it?
I could not see any property of , if I assume is not dense in .
I argue as follows. We need only prove the closure of contains
Suppose then not in the closure of , then there exists a , such that for any , the ball . Since is countable, we see This, by definition, means (the closure of ).
Here, I could only prove , but not
Would you help me out? Thank you.
Another way to define being dense-in-itself to to say that the set contains no isolated points.
Thus is dense-in-itself?
Can be dense-in-itself?
I HATE THESE TRICKY PROOFS.
Notation: where .
Note that if then contains a point because has no isolated points.
But is uncountable so
So if then such that so that AND .
Can you see why I HATE this?
There is a sequence of points in such that .
There is a sequence of points in such that .
A friend of mine came up with this proof having to do with perfect sets. By definiton, a set E is perfect if E=E', the set of limit points of E. A set is perfect if it is closed and has no isolated points. And if a set is perfect, it is uncountable. This Wikipedia article has some background:
Derived set (mathematics) - Wikipedia, the free encyclopedia
You can also find a proof that a perfect set is uncountable on the web.
Ok. Suppose that is not dense in . Then there exists and such that is empty. So is empty. This means that in every point of is also a point of E. So is a subset of E. But is closed, and since it is a subset of E, contains no isolated points. So is perfect, and therefore uncountable. This is a contradiction, so is dense in .
- Hollywood