Example verification: $E$ and $E'$ might not have the same limit points

Question

Let $E'$ be the set of all limit points of a set $E$.

Do $E$ and $E'$ always have the same limit points?

Proof: $E''\subset E'$ because $E'$ is closed.

But inclusion $E'\subset E''$ is false. We can take $\mathbb{R}^1$ with metric $\rho(x,y)=|x-y|$ and set $E=\{1/n: n\in \mathbb{N}\}$. Then $E'=\{0\}$ but $E''=\varnothing$.

Is my example true?

That is the exact counterexample I thought of. Yeah, looks good. — Plutoro, Jun 27 '15 at 19:58
Yes, this is fine. And it can be extended to get $E''$ different from both $E'$ and $E$, and so on: see this question and its answers. — Brian M. Scott, Jun 27 '15 at 20:01

score 1 · Answer 1 · answered Sep 18 '15 at 20:39

1

Your counterexample is about as good as it gets.

Interestingly, one can investigate the behaviour of iterating the operation $'$. A key phrase in this direction is Cantor-Bendixson rank.

answered Sep 18 '15 at 20:39

Lord_Farin

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Example verification: $E$ and $E'$ might not have the same limit points

1 Answers1