# Baby rudin condensation points

1. Jun 11, 2012

### 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?

2. Jun 11, 2012

### Mandlebra

3. Jun 11, 2012

### jecharla

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

4. Jun 11, 2012

### 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.

Last edited: Jun 12, 2012
5. Jun 11, 2012

### DonAntonio

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

6. Jun 12, 2012

Thanks guys!