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Condensation points in a separable metric space and the Cantor-Bendixon Theorem

  1. Apr 30, 2010 #1
    EDIT: I figured out my error, so don't worry about reading through all of this unless you find it an interesting problem

    1. The problem statement, all variables and given/known data
    This is Baby Rudin's exercise 2.27:

    [PLAIN]http://img63.imageshack.us/img63/584/fool.png [Broken]

    Instead of proving for R^k, I did it for an arbitrary separable metric space X, as outlined by professor George Bergman in his exercises to supplement Baby Rudin (http://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.ps).

    Here is what Bergman specifically says:
    [PLAIN]http://img442.imageshack.us/img442/9423/fooc.png [Broken]

    I don't so much want a solution as I do an answer as to whether I messed up a step somewhere, since my result implies a more general form of the Cantor-Bendixon Theorem; namely, any subset of a separable metric space can be partitioned into two disjoint sets, one of which is perfect and the other at most countable. I have always seen it stated only for closed subsets of complete separable spaces, but I never used the fact that the subset was closed nor that the space was complete in my answer, so I think I may be wrong.

    2. Relevant equations
    A metric space is separable if it contains an at most countable dense subset.
    The concept of a base of a metric space (defined in post below)
    A metric space is second countable if it has an at most countable base
    Lindelof's Theorem
    A set is perfect if it is closed and contains no isolated point.

    3. The attempt at a solution

    Here's the way I approached it. I'll just do an outline in this post, and then post my work for each step in the following posts. X will be a separable metric space, E will be an uncountable subset of X, and P will be the set of condensation points of E in X.

    1) Since X is separable, X is second countable (e.g., it has an at most countable base).
    2) Since X is a second countable metric space, every open cover of E has an at most countable subcover (Lindelof's Theorem).
    3) P, the set of condensation points of E, is closed.
    4) Every uncountable subset of a separable metric space has uncountably many condensation points.
    5) P has no isolated point.
    6) P is perfect.
    7) If E is uncountable, then E\P, the set of all non-condensation points of E, is at most countable.
    8) Hence, E is the union of the disjoint sets P and E\P, where P is perfect and E\P is at most countable. (whoops... messed up here)
    9) Therefore, every subset of a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 30, 2010 #2
    1) Since X is separable, X is second countable (e.g., it has an at most countable base).

    Here's the definition of a base:

    [PLAIN]http://img530.imageshack.us/img530/5992/basezy.png [Broken]

    Here's my proof of this result (I'll post an image instead of re-writing the LaTeX source, since I use a lot of user-defined commands in my LaTeX code):

    [PLAIN]http://img11.imageshack.us/img11/4599/foozc.png [Broken]

    Note I also proved a second countable metric space is separable, but didn't post that part since it's not relevant here.
     
    Last edited by a moderator: May 4, 2017
  4. Apr 30, 2010 #3
    2) Since X is a second countable metric space, every open cover of one of E has an at most countable subcover (Lindelof's Theorem).

    [PLAIN]http://img443.imageshack.us/img443/7428/fooy.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Apr 30, 2010 #4
    3) P, the set of condensation points of E, is closed.

    [PLAIN]http://img511.imageshack.us/img511/7899/foot.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Apr 30, 2010 #5
    4) Every uncountable subset of a separable metric space has uncountably many condensation points.

    [PLAIN]http://img219.imageshack.us/img219/9003/fooa.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Apr 30, 2010 #6
    5) P has no isolated point.
    6) P is perfect

    [PLAIN]http://img199.imageshack.us/img199/9003/fooa.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Apr 30, 2010 #7
    7) If E is uncountable, then E\P, the set of all non-condensation points of E, is at most countable.

    [PLAIN]http://img718.imageshack.us/img718/9840/foob.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Apr 30, 2010 #8
    8) Hence, E is the union of the disjoint sets P and E\P, where P is perfect and E\P is at most countable, from 7).
     
  10. Apr 30, 2010 #9
    9) Therefore, every subset of a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.

    [PLAIN]http://img28.imageshack.us/img28/7880/foohks.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  11. Apr 30, 2010 #10
    I ask because it seems too general, considering Rudin asks the reader to prove the result #9) under the constraint that the subset is closed. Is my result wrong, or did Rudin just want the reader to prove an easier version? Knowing Rudin and his problems, I'm leaning towards the former. Thanks in advance!
     
  12. Apr 30, 2010 #11
    One more note on notation:

    [tex]B_r(x)[/tex] means the open ball of radius r, centered at point x.

    e.g.,

    [tex]B_r(x) = \{y \in X\ | \ d(y,x) < r\}[/tex]
     
  13. Apr 30, 2010 #12
    Eh, never mind. I made a stupid mistake in part 8). P is contained in the set of limit points of E, which is contained in E when E is closed. LOL @ me being tripped up by the easiest part of the problem. At least my intuition that Rudin wouldn't let the reader get away with proving a lesser version of a theorem was dead on.
     
    Last edited: Apr 30, 2010
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