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.