For any $n \in \Bbb N$ , does there exist $A \subset \Bbb R$ such that $A^1,A^2,\ldots,A^{n-1}$ are non-empty and $A^n = \emptyset$?

Question

In a topological space $X$ , $a$ is defined to be a condensation point of a set $A$ in $X$ if and only if each neighborhood of $a$ meets $A$ in uncountably many points.

Let $A^c$ denote the set of all condensation points of A. Further denote $A^1 = A^c$ , $A^2 = (A^1)^c$ , $A^3 = (A^2)^c \text{ , }\ldots \text{ , } A^{j+1} = (A^j)^c \text{ , } \forall j \in \Bbb N$ . Then clearly, $A^1 \supset A^2 \supset \ldots \supset A^j \supset A^{j+1} \supset \ldots \text{ etc. }$

My question is : For any $n \in \Bbb N$ , does there exist $A \subset \Bbb R$ such that $A^1,A^2,\ldots,A^{n-1}$ are non-empty and $A^n = \emptyset$ ? (considering in the Usual topology)

Tried the problem but couldn't come up with anything.

Answer 1

First a comment

The set of condensation points of a subset of the reals maybe bigger than the original set. For example $(0,1)^c=[0,1]$. See condensation point.

Second regarding your question

Following affirmations are true:

1. The set of condensation points $A^c$ of a subset of the reals $A \subseteq \mathbb R$ is perfect.
2. Every point of a perfect set is a condensation point for that set. See theorem 13 of the link.
3. A non-empty perfect set has the cardinality of the continuum.

Therefore for $A \subseteq \mathbb R$, either $A^c$ is empty and you have $A^1=\emptyset$. Either $A^1$ is a non-empty perfect set and the sequence $A^n$ is constant non-empty for $n\ge1$.