Why Is Every Point of P a Limit Point of P?

[jecharla]

1st part of Exercise #27 is:

Define a point p in a metric space X to be a condensation point of a set E in X if every neighborhood of p contains uncountably many points of E. Suppose E is in R^k, E is uncountable and let P be the set of all condensation points of E. Prove P is perfect.

Obviously, P is closed. But I cannot figure out why every point of P is a limit point of P. There are several supposed solutions to this in the internet, but each of them only shows that if x is in P, x is a limit point of E(which is obvious since x is a condensation point of E). But the exercise asks us to prove that P is perfect, so if x is in P, x must be a limit point of P right? Could anyone offer any guidance to what such a proof would look like?

 

[Mandlebra]

Could you define perfect please?

 

[jecharla]

A set S is perfect if S is closed and if every point of S is a limit point of S.

 

[Undecided Guy]

Suppose a point x in P is isolated. Then there's an \epsilon > 0 so that B(x; \epsilon) contains no other point of P. Since x is in P, this ball contains uncountably many points of E. Note that we may write B(x; \epsilon) = \bigcup_{j \in J} B(x_j; r_j) for each r_j < \epsilon and x_j with rational coordinates where J is countable. Thus for at least some j we must have that B(x_j; r_j) contains uncountably many points of E (countable unions of countable sets are countable).

Fix this j. Since B(x_j ; r_j) = \bigcup_{q < r_j, q \in \mathbb{Q}} B(x_j; q) and this is once again a countable union, we must have again that there is some q_1 < r_j so that B(x_j; q_1) contains uncountably many points of E. Proceeding by induction, we may construct a decreasing sequence of rational numbers so that the ball centered at x_j of each of these radii contains uncountably many points of E. This shows that x_j is in P, contrary to our assumption.

Edit: thinking about it, I don't think you can easily show that \{q_n\}_{n=1}^{\infty} tends to 0 the way this is set up, which is required for the contradiction.

 

[DonAntonio]

A set S is perfect if S is closed and if every point of S is a limit point of S.

Take \,p\in P\, and let \,U:=B_\epsilon(p)\, be any open ball of positive radius around \,p\, and

let \overline{U} be its closure.

By definition, \,\overline{U}\, contains uncountable many points of \,E\, , and since it is a compact

set then for any \,\epsilon>0\, there exist only a finite number of balls of radius \,\epsilon\, covering it. This means that there can be only

at most a countable number of points in \,U\, which are not elements of \,P\, (why? Something must be argued here!)

Thus, as in U we have uncountable points of E and only countably many of them are

not in P there are left enough points from which we can form a sequence in P that converges to p.

DonAntonio

 

[jecharla]

Thanks guys!