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Define a point $p$ in a metric space $X$ to be a condensation point of a set $E\subset X$ if every neighborhood of $p$ contains uncountably many points of $E$.

Suppose $E\subset \mathbb{R}^k$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect.

1) I proved that $P$ is closed set.

2) But how to prove than any point of $P$ is a limit point of $P$?

Can anyone give a solution to 2)?

I saw a lot of links but no one of them helps to me.

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    $\begingroup$ Suppose $x_0$ were an isolated point of $P$. Then there is an $r > 0$ such that $P \cap B_r(x_0) = \{x_0\}$. But $E\cap B_\rho(x_0)$ is uncountable for every $\rho > 0$. Use that to conclude that there must be condensation points of $E$ in $B_r(x_0) \setminus B_\rho(x_0)$ for some $\rho \in (0,r)$. $\endgroup$ – Daniel Fischer Aug 6 '15 at 9:26
  • $\begingroup$ Dear, Daniel Fischer! I think about what you wrote some hours but any results. Can you help me? $\endgroup$ – ZFR Aug 6 '15 at 12:44
  • $\begingroup$ By contradiction. Let for any $\rho\in(0,1)$ we have that $(B_{r}(x_0)\backslash B_{\rho}(x_0))\cap P = \varnothing$. $\endgroup$ – ZFR Aug 6 '15 at 12:48
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    $\begingroup$ @FardadPouran Assume $x_0 \in P$ were not a condensation point of $P$. Then there is an $r > 0$ such that $B_r(x_0) \cap P$ is countable. Since $P$ is closed, $B_r(x_0) \setminus P$ is an open set, and it's non-empty since $B_r(x_0) \cap P$ is countable. And since $x_0$ is a condensation point of $E$, $(B_r(x_0)\setminus P) \cap E$ is uncountable. Conclude that it must contain a condensation point of $E$, which hence belongs to $P$. Contradiction. $\endgroup$ – Daniel Fischer Oct 15 '15 at 18:54
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    $\begingroup$ @kishlaya Condensation points are limit points, so that holds for all closed sets. $\endgroup$ – Daniel Fischer Oct 5 '17 at 9:50
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Here is a nice proof.

I'll prove that any point of $P$ is a limit point of $P$.

Proof: Let $z$ is not limit point of $P$ then $\exists\varepsilon>0$ such that $N'_{\varepsilon}(z)\cap P=\varnothing$. Also we can change $\varepsilon$ such that $\bar{N}'_{\varepsilon}(z)\cap P=\varnothing$.

Let $A_n=\{x\in R^k: 1/n\leqslant d(x,z)\leqslant \varepsilon\}.$ All $A_n$ is closed and bounded $\Rightarrow$ $A_n$ are compact for any $n\in \mathbb{N}$. It's easy to check that $\bar{N}'_{\varepsilon}(z)=\cup_{n\geqslant 1}A_n$.

We'll built an open cover for $A_n$. For any $x\in A_n$ we have $x\in \bar{N}'_{\varepsilon}(z)$ but $x\notin P$. Hence $\exists \varepsilon_x$ such that $N_{\varepsilon_x}(x)$ contains at most countably many points of $E$. So $A_n$ is compact then $\{N_{\varepsilon_i}(x_i)\},$ $1\leqslant i\leqslant m_n$ is a finite subcover of $A_n$, where $\varepsilon_i=\varepsilon_{x_i}$. So $A_n$ contains at most countable points of $E$. Using that $\bar{N}'_{\varepsilon}(z)=\cup_{n\geqslant 1}A_n$ we got that $\bar{N}'_{\varepsilon}(z)$ contains countably many points of $E$ which is absurd because $z\in P$.

Q.E.D.

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