# rudin pma exercise 2.27

Please check if my proof is right or wrong. Any advice is welcome. Moreover how can I check my proofs right or wrong when I solve some problem by myself?

(1) $P$ is closed

Let's assume point $z$ is a limit point of $P$ but $z$ is not a point of $P$. Since $z$ is a limit point, for all $r$ $p_i \in N_r(z)$ satisfies where some points $p_i\in P$. Let $\mathcal{E}<r-d(z,p_i)$ then $N_\mathcal{E}(p_i)$ contain uncountably many points of $E$ and $N_\mathcal{E}(p_i) \subset N_r(z)$ so $N_r(z)$ contain uncountably many points of $E$. $P$ is a set of all condensation point of $E$. $z$ should be a point of $P$

(2) every point of $P$ is a limit point

Assume every point of $P$ is not a limit point i.e. there exists r that $N_r(p_i)$ doesn't contain points in $P$ (other than $p_i$). By definition of condensation set $N_r(p_i)$ contain uncountably many point of $E$. By assumption, all points in $N_r(p_i)$ that neighborhood of that points contain countably many point of $E$ (if there exists a point that neighborhood is uncountably many point of $E$, this point should be in set $P$). Then $N_r(p_i) \subset \bigcup_x^\inf N_{r'}(x)$ Union of countable set is countable so $N_r(p_i)$ should be countable which contradicts to definition of condensation set.

by (1) and (2) $P$ is perfect.

(3) $P^c \cap E$ is at most countable.

Following hint and previous exercises of pma 2.22,2.23, $R^k$ is seperable (2.22) and every seperable metric space has a countable base (2.23). So we can think ${V_n}$ as hint. $P$ is nonempty perfect set in $R^k$ so $P$ is uncountable (pma theorem 2.43). It is obvious that $P \subset W^c$ (if $x \in P$ then $x \in W$, it contradicts that $P$ contains uncountably many points of $E$). Conversly suppose $x \in W^c$. Then x is a point of $V_n$ where $E \cap V_n$ is uncountable. Since $V_n$ is a base and for any neighborhood $N(x)$ of $x$, $x \in V_n \subset N(x)$. Thus $E \cap N(x)$ is also uncountable that means $x \in P$. So $W^c \subset P$. Thus $P=W^c$

Let $V_n$ be the countable base you know exists.
Define as hinted: $W = \cup \{V_n: V_n \cap E \text{ at most countable }\}$.
The claim made is $P = W^c$. So suppose $x \in P$, this means that $x$ is a condensation point of $E$, by definition. This implies that for any $V_n$ that contains $x$, $V_n \cap E$ is uncountable, so $x$ cannot be in this union $W$. So $P \subseteq W^c$.
Suppose that $x \notin P$. Then there is some neighbourhood $O$ of $x$ that intersects $E$ in a set that is at most countable. For some $n$, $x \in V_n \subset O$ and this $V_n$ witnesses that $x \in W$. So $P^c \subset W$ or equivalently $W^c \subset P$, hence equality.
$P$ is then closed as the complement of the open set $W$. $P$ is perfect as it is closed, and no point of $P$ is isolated in $P$, as you showed.
Then all points of $E$ that are not condensation points is just $E\setminus P =E\cap P^c = E \cap W$. The latter set is countable as all intersections with the $V_n$ that constitute $W$ are countable, and we have countably many of those intersections.