How is the set of condensation points of an uncountable subset of $\Bbb{R}^k$ perfect?

Question

My textbook defines a perfect set $E$ to be a closed set such that every point in $E$ is a limit point of $E$.

It also asks me to prove the following:

Suppose $E\subset \Bbb{R^k}$, E is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect.

I wonder why that is. If $E=(0,1)\subset\Bbb{R^1}$, then $P=\{0,1\}$. How is $P$ perfect? It is closed, but neither $1$ nor $0$ are limit points of the set.

Thanks in advance!

Finite sets have no limit points. So $A={1}$ is a set where $1\in A$ but 1 is not a limit point of $A$. — gradstudent, Feb 23 '14 at 03:41

score 2 · Answer 1 · answered Feb 23 '14 at 03:36

2

The set of condensation points of $E = (0,1)$ is not $\{0,1\}$, but the entire closed interval $[0,1]$. $x$ is a "condensation point" if every open ball around $x$ contains uncountably many points in $E$--this can be the case whether or not $x$ itself is a member of $E$.

Re edit: $\{1\}$ is not a perfect set in $\mathbb{R}^1$. $1$ is not a limit point of $\{1\}$, so the set is not perfect.

answered Feb 23 '14 at 03:36

Caleb Stanford

45,895

Ah. Thanks for your help. – Feb 23 '14 at 03:39

How is the set of condensation points of an uncountable subset of $\Bbb{R}^k$ perfect?

1 Answers1