TITLE: Infinite removal of isolated points from a subset of $\mathbb{R}$.
QUESTION [6 upvotes]: Assume we have subset of $\mathbb{R}$. Then, at every step, we remove all isolated points from what is remained from initial subset. We stop when there is nothing to remove - so current set is either empty or there is no isolated points. Is there a subset such that we never stop?
Intuition tells me what we should stop after step one - however it is not ture.
I can provide an example when we stop after two steps. We should take $\{0\}\cup\{{\frac{1}{n}\}}_{n \in{\mathbb{N}}}$. In this set every point except $0$ is isolated, so $0$ will survive after first iteration.
I can not provide an example more steps and I do now know if it is possible to constuct a subset for infinite number of iterations. Any ideas?

REPLY [2 votes]: Nice question. I don't know the general answer yet, sadly, but can offer one for a three-step isolation.
Take your set, $$\{0\}\cup \left\{\frac 1n;n\in \mathbb N\right\}$$
Now, for each $n\in\mathbb N$, take a sequence $a_k^{(n)}$ which converges to $\frac 1n$ and make a set with all elements of the added sequence, along with elemnents from your original set.
In this new set, the first step will remove all sequences $a^{(n)}$. The second will remove $\frac1n$ values, and the third will remove $0$.

Iteratively repeating this process will allow you to construct a set $A_k$ which will require $k$ removals of isolated points to clear. I am not sure what would happen if you repeated the construction process infinity-times.