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.

Thank you for your help.

  • 1
    $\begingroup$ 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 $R^k$. $\endgroup$ – hot_queen Dec 28 '13 at 6:20
  • 1
    $\begingroup$ 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. $\endgroup$ – hot_queen Dec 28 '13 at 6:29
  • $\begingroup$ @hot_queen : Error fixed. Please discuss in detail and write down the answer. $\endgroup$ – Dutta Dec 28 '13 at 10:23
  • $\begingroup$ Left as an exercise: abstrusegoose.com/12 $\endgroup$ – dspyz May 8 '14 at 20:01
  • 2
    $\begingroup$ "countably many" should be "uncountably many" in the definition of "condensation point", I think. $\endgroup$ – 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 $x \in E \setminus P$.

  • $\begingroup$ Long while I am out of touch from topology. But thanks for your interest. $\endgroup$ – Dutta Feb 12 '15 at 7:25

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.