# Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$

Suppose that $E\subset \Bbb R^k$, $E$ is uncountable and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that atmost countably many points of $E$ are not in $P$.

Attempt:

To show that $P$ is perfect I show that $P$ is closed and every point of $P$ is a limit point of $P$.

Let $p$ be a limit point of $P$,then $B(p,r)\cap P$ will contain infinitely many points of $P$. Now if $a\in B(p,r)\cap P$ then $a$ is a condensation point and $B(p,r)$ is a neighbourhood of $a$ and hence $B(p,r)$ contains uncountably many points of $E$ and hence $p\in P$.Hence $P$ is closed.

Now to show that every point of $P$ is a limit point of $P$.

let $a\in P$ to show that $a$ is a limit point of $P$,consider $B(a,r) ;r>0$

Then $B(a,r)$ will contain uncountably many points of $E$.Choose $b\in B(a,r)$ then $B(a,r)$ is a neighbourhood of $b$ and also contains uncountably many points of $E$ and hence $b\in P\implies b\in B(a,r)\cap P\implies a$ is a limit point of $P$.

Problem

Unable to show that atmost countably points of $E$ are not in $P$ i.e. to show that $|E\setminus P|$ is countable.

Are the above proofs correct? How to solve the problem?Any help.

Warning: this proof is false see Mechanandroid's comment below and the definition of a condensation point.

One has

$$E\setminus P = \bigcup_{n=1}^\infty \big\{x\in E: d(x, E\setminus\{x\})>\frac{1}{n}\big\} := \bigcup_{n=1}^\infty E_n$$ Each $E_n$ is countable because the distance between two arbitrary points of $E_n$ is larger than $\frac{1}{n}$. As a consequence there can be only a finite number of elements of $E_n$ in any bounded region of ${\mathbb R}^k$.

It follows that $E\setminus P$ is countable.

• Why must $B(x,r_x)$ be disjoint? You only know that $E \cap B(x,r_x)$ are disjoint. – mechanodroid Sep 16 '17 at 11:21
• You are right, they don't need to be disjoint. I'm thinking... – Gribouillis Sep 16 '17 at 11:23
• @mechanodroid I edited the answer. – Gribouillis Sep 16 '17 at 14:28
• I think you made the same mistake I did in my answer: $x\in P$ implies there exists $r>0$ such that $B(x,r)\cap E$ is countable, not $B(x,r)\cap E = \{x\}$. So you can have $x\in E\setminus P$ but $d(x,E\setminus \{x\})=0$. – mechanodroid Sep 16 '17 at 14:39
• @mechanodroid All right, forget it :) Thank you. – Gribouillis Sep 16 '17 at 14:43