enter image description here

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 $P=W^c$


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.