Condensation Points and Perfect Sets (Proof Verification)

Question

I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt:

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

My Proof:

Suppose a limit point of $P$ is not in $P$, and call this point $z$. Then, every neighborhood of radius $r$, $N_r(z)$, contains a point $p \in P$. But every neighborhood of $p$ contains uncountably many points of $E$, including $N_{r-d(z,p)}(p)$. Let $a \in N_{r-d(z,p)}(p)$, and we can see that:

$d(z,a) \le d(z,p) + d(p,a) < d(z,p) + (r - d(z,p)) = r$

So $a \in N_r(z)$, and $N_{r-d(z,p)}(p) \subset N_r(z)$. Since $N_r(z)$ has a subset with uncountably many points of $E$, it has uncountably many points of $E$. This contradicts our assumption that $z$ was not in $P$, so all limit points of $P$ must be in $P$, so $P$ is closed.

Now suppose $z$ is an isolated point of $P$. Then, for some $r$, $N_r(z)$ has uncountably many elements of $E$ but contains no other elements of $P$. However, that implies that for any point $x$ s.t. $d(z,x) = \frac{r}{2}$, $N_{\frac{3r}{2}}(x)$ also contains uncountably many elements of $E$, so $x \in P$, which contradicts our assumption that $N_r(z)$ contains no elements of $P$ other than $z$. Therefore, there are no isolated points in $P$.

Since $P$ is closed and has no isolated points, $P$ is perfect.

Answer 1

Your proof that $P$ is closed is correct but unnecessarily clumsy. It could be condensed as follows.

Suppose that $z$ is a limit point of $P$. Then for any $r>0$ there is a $p\in N_r(z)\cap P$. $N_r(z)$ is an open nbhd of $p$, and $p\in P$, so $N_r(z)\cap E$ is uncountable. Thus, every open nbhd of $z$ contains uncountable many points of $E$, and therefore $z\in P$, and $P$ is closed.

Remember, open nbhds of points in $\Bbb R^k$ aren’t limited to open balls.

The second part, however, doesn’t work: you’ve shown that one open nbhd of $x$ contains uncountably many points of $E$, but in order to show that $x\in P$, you have to show that every open nbhd of $x$ contains uncountably many points of $E$. Here’s an argument that does work:

Let $x\in P$, and suppose that $x$ has an open nbhd $V$ such that $V\cap P=\{x\}$. Let $U=V\setminus\{x\}$, and note that $U$ is open. Let $\mathscr{B}$ be a countable base for $\Bbb R^k$. If $y\in U$, then $u\notin P$, so $y$ has an open nbhd $W_y$ that contains only countably many points of $E$, and there is a $B_y\in\mathscr{B}$ such that $y\in B_y\subseteq W_y$; clearly $B_y\cap E$ is countable. Let $\mathscr{B}_0=\{B_y:y\in U\}$; then $\mathscr{B}_0$ is countable, and $U=\bigcup\mathscr{B}_0$, so $$U\cap E=\left(\bigcup\mathscr{B}_0\right)\cap E=\bigcup\{B_y\cap E:y\in U\}$$ is a countable union of countable sets and is therefore also countable. But then $V\cap E$ is countable, contradicting the fact that $x\in P$, and it follows that $x$ is not isolated.