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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.

Martin Sleziak
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RFZ
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    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 $\rho \in (0,r)$. – Daniel Fischer Aug 06 '15 at 09:26
  • Dear, Daniel Fischer! I think about what you wrote some hours but any results. Can you help me? – RFZ Aug 06 '15 at 12:44
  • By contradiction. Let for any $\rho\in(0,1)$ we have that $(B_{r}(x_0)\backslash B_{\rho}(x_0))\cap P = \varnothing$. – RFZ Aug 06 '15 at 12:48
  • Related posts: http://math.stackexchange.com/questions/345841/baby-rudin-to-show-the-set-of-all-condensation-points-of-a-set-in-euclidean-sp and http://math.stackexchange.com/questions/240593/how-to-prove-that-the-set-of-condensation-points-of-an-uncountable-subset-of-the – Martin Sleziak Aug 06 '15 at 13:36
  • @DanielFischer, I got your hint. But how can we go beyond perfectness and prove that each point of $P$ is a condensation point of $P$ ? – Fardad Pouran Oct 15 '15 at 18:43
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    @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. – Daniel Fischer Oct 15 '15 at 18:54
  • Thank you, I successfully proved it via Lindolf property – Fardad Pouran Oct 16 '15 at 18:12
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    @kishlaya Condensation points are limit points, so that holds for all closed sets. – Daniel Fischer Oct 05 '17 at 09:50

1 Answers1

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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.

RFZ
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