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Baby rudin condensation points

  1. Jun 11, 2012 #1
    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. jcsd
  3. Jun 11, 2012 #2
    Could you define perfect please?
  4. Jun 11, 2012 #3
    A set S is perfect if S is closed and if every point of S is a limit point of S.
  5. Jun 11, 2012 #4
    Suppose a point x in P is isolated. Then there's an [itex] \epsilon > 0 [/itex] so that [itex]B(x; \epsilon)[/itex] contains no other point of P. Since x is in P, this ball contains uncountably many points of E. Note that we may write [itex]B(x; \epsilon) = \bigcup_{j \in J} B(x_j; r_j) [/itex] for each [itex]r_j < \epsilon[/itex] and [itex] x_j[/itex] with rational coordinates where J is countable. Thus for at least some j we must have that [itex]B(x_j; r_j) [/itex] contains uncountably many points of E (countable unions of countable sets are countable).

    Fix this j. Since [itex]B(x_j ; r_j) = \bigcup_{q < r_j, q \in \mathbb{Q}} B(x_j; q) [/itex] and this is once again a countable union, we must have again that there is some [itex] q_1 < r_j [/itex] so that [itex]B(x_j; q_1)[/itex] 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 [itex]\{q_n\}_{n=1}^{\infty} [/itex] tends to 0 the way this is set up, which is required for the contradiction.
    Last edited: Jun 12, 2012
  6. Jun 11, 2012 #5

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

    let [itex]\overline{U}[/itex] be its closure.

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

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

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

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

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

  7. Jun 12, 2012 #6
    Thanks guys!
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