4
$\begingroup$

Suppose that $E\subset \Bbb R^k$, $E$ is uncountable and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that atmost countably many points of $E$ are not in $P$.

Attempt:

To show that $P$ is perfect I show that $P$ is closed and every point of $P$ is a limit point of $P$.

Let $p$ be a limit point of $P$,then $B(p,r)\cap P$ will contain infinitely many points of $P$. Now if $a\in B(p,r)\cap P $ then $a$ is a condensation point and $B(p,r)$ is a neighbourhood of $a$ and hence $B(p,r)$ contains uncountably many points of $E$ and hence $p\in P$.Hence $P$ is closed.

Now to show that every point of $P$ is a limit point of $P$.

let $a\in P$ to show that $a$ is a limit point of $P$,consider $B(a,r) ;r>0$

Then $B(a,r)$ will contain uncountably many points of $E$.Choose $b\in B(a,r)$ then $B(a,r)$ is a neighbourhood of $b$ and also contains uncountably many points of $E$ and hence $b\in P\implies b\in B(a,r)\cap P\implies a$ is a limit point of $P$.

Problem

Unable to show that atmost countably points of $E$ are not in $P$ i.e. to show that $|E\setminus P|$ is countable.

Are the above proofs correct? How to solve the problem?Any help.

$\endgroup$
0
$\begingroup$

Warning: this proof is false see Mechanandroid's comment below and the definition of a condensation point.

One has

$$ E\setminus P = \bigcup_{n=1}^\infty \big\{x\in E: d(x, E\setminus\{x\})>\frac{1}{n}\big\} := \bigcup_{n=1}^\infty E_n $$ Each $E_n$ is countable because the distance between two arbitrary points of $E_n$ is larger than $\frac{1}{n}$. As a consequence there can be only a finite number of elements of $E_n$ in any bounded region of ${\mathbb R}^k$.

It follows that $E\setminus P$ is countable.

$\endgroup$
  • $\begingroup$ Why must $B(x,r_x)$ be disjoint? You only know that $E \cap B(x,r_x)$ are disjoint. $\endgroup$ – mechanodroid Sep 16 '17 at 11:21
  • $\begingroup$ You are right, they don't need to be disjoint. I'm thinking... $\endgroup$ – Gribouillis Sep 16 '17 at 11:23
  • $\begingroup$ @mechanodroid I edited the answer. $\endgroup$ – Gribouillis Sep 16 '17 at 14:28
  • $\begingroup$ I think you made the same mistake I did in my answer: $x\in P$ implies there exists $r>0$ such that $B(x,r)\cap E$ is countable, not $B(x,r)\cap E = \{x\}$. So you can have $x\in E\setminus P$ but $d(x,E\setminus \{x\})=0$. $\endgroup$ – mechanodroid Sep 16 '17 at 14:39
  • $\begingroup$ @mechanodroid All right, forget it :) Thank you. $\endgroup$ – Gribouillis Sep 16 '17 at 14:43

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.