Here is the theorem to be proved:

**Suppose $E subset mathbb{R}^k$, $E$ is uncountable, and $P$ is the set of condensation points of $E$. Prove that $P$ is perfect and that there are at most countably many points in $E$ that are not in $P$**.

I tried to prove it as follows:

Let $pin mathbb{R}^k$ be a limit point of $P$, then for some $delta gt 0, N_delta(p)$ contains $yne x$ such that $yin P$. Let $r=frac{1}{2} min (d(p,y), delta-d(p,y))$. We have $N_r(y)subset N_delta (p)implies N_r(y) $ contains uncountable many points of $E$ (as $y$ is condensation point)$implies N_delta (p)$ contains uncountable many points of $Eimplies p$ is a condensation point of $E$. So $pin P$. Hence $P$ is closed. $tag{1}$

Let $pin P$ be an isolated point of $P$. $existsdelta gt 0: N_delta (p)cap P={p}$. Since $pin P$, we have that $N_delta (p)cap E$ is uncountable.

Suppose that $N_delta (p)$ does not contain any other condensation point of $E$ except $p$.

Then for every $yin N_delta (p)-{p}$, and for $r= 0.5 min(d(p,y),delta – d(p,y)), N_r (y)$ is at most countable for every $yin N_delta (p)-{p}$.

We have $N_delta (p)subset (cup_{yin N_delta (p)-{p}} N_r(y))cup{p}implies Ecap N_delta (p)subset (Ecap {p})cap_{yin N_delta (p)-{p}} (Ecap N_r(y)subset Ecap N_r(y)$ for some $yin N_delta (p)-{p}implies N_delta(p)$ is at most countable, which is a contradiction. Therefore, all points of $P$ are limit points of $P$. $tag{2}$ **By $(1)$ and $(2), P$ is perfect.**

I got stuck at proving the second part of this question.

I tried to use hint provided in the exercise: “Let ${V_n}$ be a countable base of $mathbb R^k$, let $W$ be the union of those $V_n$ for which $Ecap V_n$ is at most countable and show that $P=W^c$.”

I understand that since $mathbb R^k$ is separable, it must have a countable base. I don’t know how to proceed further using the hint. I also don’t understand why such $W$ exists. Can existence of such $W$ be proven?

Please help. Thanks.