I'm trying to prove the following:
For every positive integer $n,$ there is $A\subset{\mathbb{R}}$ such that $A,A^{1},A^{2},…,A^{n-1}$ are nonempty and $A^{n}=\emptyset.$ (The result can be extended to countable ordinals. Let $A^{\alpha}=(A^{\alpha-1})'$ if $\alpha$ is a nonlimit ordinal, $A^{\alpha}=(\bigcap_{\beta<\alpha}A^{\beta})'$ if $\alpha$ is an ordinal limit, and show that for any $\alpha<\omega_{1},$ a set $A$ can be finding such that $A^{\alpha}=\emptyset$ and $A^{\beta}\neq{\emptyset}$ for $\beta<\alpha$).
Here $A^{1}=(A)'$ is the set accumulation points of $A.$ Then $A^{2}=(A^{1})^{'}$ and so on.
$\omega_{1}$ is the first uncountable ordinal.
I am stuck with this. For the case $n=1$ is enough consider a non-constant convergent sequence, but for the other cases, I have only found cases in which the set of accumulation points is uncountable ( intervals ) with no empty set accumulation points or countable sets with empty set as accumulation point. Building sets with finite accumulation points is no problem, but asked the required property is strange.
Any kind of help is thanked in advanced.
Best Answer
For $n = 1:\{0\}$
For $n = 2:\{ 1/n : n\in\Bbb N\}\cup\{0\}$
For $n = 3:\{ 1/n + 1/m : n,m\in\Bbb N\}\cup\{ 1/n : n\in\Bbb N\}\cup\{0\}$
The idea is to take the sequence $(1/n)_n$ and for
each point $1/n$ create a sequence with limit of $1/n$.
This construction can be extended to any finite $n$.