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.
Best Answer
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.