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 $A^n = \emptyset$?

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.

• For spaces maybe more exotic than $\mathbb{R}^n$ it also makes sense to ask the same question for $\kappa$-condensation points, where $\kappa$ is a cardinal, i.e. points with $>\kappa$ points in every neighbourhood (so that a condensation point is an $\aleph_0$-condensation point). – Qfwfq Aug 14 '18 at 17:53
• Question already asked on MSE - math.stackexchange.com/questions/2882656/… – mathcounterexamples.net Aug 14 '18 at 18:08

If the space $A$ is hereditarily Lindelof (as is in the case of $A\subset \mathbb R$), then the set $A^1$ of condensation points of $A$ coincides with $A^2$. Indeed, the complement $A\setminus A^1$ is locally countable and being hereditarily Lindelof, is countable. So, for any point $a\in A^1$ any any neighborhood $O_a$ of $a$ in $A$ the cardinalities of $O_a$ and $O_a\cap A^1$ coincide, which means that $a$ is a condensation point of $A^1$ and hence $A^2=A^1$.