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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?

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Note that condensation points of $E$ may not be elements of $E$. Removing these extraneous points from the perfect set $P$ may result in an imperfect set. –  Arthur Fischer Aug 11 '12 at 17:59
    
@Arthur Oh right.. This becomes a lame question ;(... –  Katlus Aug 11 '12 at 18:01
    
It is a good question for example for me because I am trying to learn sth..)) –  Seyhmus Güngören Aug 11 '12 at 18:08

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up vote 1 down vote accepted

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 $(0,a) \subseteq (0,1) \setminus P$!

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