# Complement of set of all condensation point for an uncountable set of reals is at most countable.

Perfect Set: A set $E \subset X$ is said to be perfect if $E$ is closed in the metric space $(X,d)$ and every point of $E$ is a limit point of $E$.

Condensation Point : A point $p \in X$ is said to be a condensation point of $E \subset X$ if every nbd of $p$ contains countably many points of $E$.

Problem: Suppose $E \subset \mathbb{R}^k$ is an uncountable set and let $P$ be the set of all condensation points of the set $E$. Prove that $P$ is perfect and at most countably many points of $E$ will not belong to $P$.

My attempt:

I can prove that $P$ is perfect. Let $p$ be a limit point of $P$. Then any deleted nbd of $p$ say $N'(p,\delta)$ will contain a point say $q \in P$. Again $N(p,\delta)$ will be a nbd of $q$ which is a condensation point of $E$ and so $N(p,\delta)$ will contain uncountably many points of $E$. So any nbd of $p$ will contain uncountably many points of $E$. So $p \in P$ and $P$ is closed.

Let $p \in P$. $N(p,\delta)$ will contain uncountably many points of $E$. Let $q \in N'(p,\delta)$. Thus $q$ is a condensation point of $E$ and hence $q \in P$. That gives $p$ is a limit point of $P$. So $P$ is perfect.

I have no idea for the next part i.e. $P$ will not contain at most countbly many points of $E$. Please discribe it in a easy manner.

• You should replace "finitely" by "countably" in the statement of your problem. And perfect generally means complete and without isolated points so you should assume furthermore that E is closed in $R^k$. – hot_queen Dec 28 '13 at 6:20
• And to get rid of non condensations pts you can union up ctbly many basic open balls in which E is ctble. This only deletes ctbly many pts of E. – hot_queen Dec 28 '13 at 6:29
• @hot_queen : Error fixed. Please discuss in detail and write down the answer. – Dutta Dec 28 '13 at 10:23
• Left as an exercise: abstrusegoose.com/12 – dspyz May 8 '14 at 20:01
• "countably many" should be "uncountably many" in the definition of "condensation point", I think. – Trevor Wilson May 8 '16 at 3:15

First, observe that any collection of disjoint non-empty open sets in $\mathbb{R}^n$ is at most countable; each contains a rational point and no two can contain the same point.
Now, suppose $x \in E \setminus P$. Then by definition, $x$ has a neighbourhood containing finitely many points of $E$. Hence we can define $\epsilon_{x} = \underset{y \in E \setminus \{x\}}{\inf} \{d(x,y)\}$ and know that $\epsilon_{x} > 0$, since either $x$ has a neighbourhood containing finitely many point of $E$ other than $x$, and $\epsilon_{x}$ is the distance to the nearest point, or $x$ has no such neighbourhood (e.g. $E$ is the unit circle together with $(0,0)$), but does have a neighbourhood containing just $x$, so again $\epsilon_{x}>0$.
Now, consider the collection of non-empty open balls $B_x=N(x,\epsilon_{x}/2)$. These are disjoint, since if $B_x \cap B_y \neq \emptyset$ then $d(x,y) < 1/2 (\epsilon_{x}+\epsilon_{y}) \leq 1/2(d(x,y) + d(y,x))$. By the opening comment, there are at most countably many of them, hence countably many $x \in E \setminus P$.