Cantor Bendixson Theorem: Every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.
This definition differs a bit from that in wikipedia.
I have proved that 'If $X$ is a separable metric space and $E$ is a uncountable subset and $P$ is the set of all condensation points of $E$, then $P$ is perfect and $P^c \cap E$ is at most countable'.
Then, you can see that 'every uncountable set in a separable metric space is the union of a nonempty perfect set and a set which is at most countable, and sets are disjoint' (Since $E= P\cup (P^c \cap E)$)
Here, i didn't use the condition 'closed' at all!
Where did i go wrong?
Best Answer
I'll just put my comment into an answer.
The problem comes in the assertion that $E = P \cup ( P^c \cap E)$. In your construction, you have that $P$ is the set of all condensation points of $E$, but for arbitrary $E$ not all condensation points of $E$ belong to $E$. (As a simple example, note that the set of all condensation points of the open interval $(0,1)$ in the real line is the closed interval $[0,1]$.)
If one were to, instead, consider $(P \cap E ) \cup ( P^c \cap E)$ we quickly run into the problem that $P \cap E$ is not perfect -- as the above example would show.
As a conclusion of sorts, one can easily show that $(0,1)$ cannot be represented as the union of a perfect set and a countable set in the real line: Suppose that $P \subseteq (0,1)$ is any perfect set. As $P$ is bounded(-below) and closed, it must have a minimum element, $a$, and $0 < a < 1$. As we all know, $(0,a)$ is uncountable, as we're done as $(0,a) \subseteq (0,1) \setminus P$!