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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?

Hedgehog
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    For any ordinal $\alpha\lt\omega_1$ there are sets of reals requiring $\alpha$ iterations of "removing isolated points". I believe this was Cantor's motivation for introducing transfinite ordinal numbers to mathematics. Look up "Cantor-Bendixson derivative". – bof Oct 28 '14 at 13:55
  • @bof I am sorry I can not now appreciate your answer - I do not know ordinals. I will learn you as soon as I can, but at this moment I can not benefit or your comment and I even do not understand the relation of the comment with the answer. – Hedgehog Oct 28 '14 at 14:00
  • The answer to your question is that, for some subsets of $\mathbb R$, the process of removing isolated points can go on for infinitely many steps, and then infinitely many more steps, and so on. I did not show you how to construct examples; that would be an Answer rather than a Comment. However I gave you some useful Keywords. They are Cantor-Bendixson derivative. – bof Oct 28 '14 at 14:07
  • @bof, thanks a lof for you comment and for your patience. I will learn everything I can and check everything for sure, also I am trying to construct the example with the help of $5xum$ answer. thanks again. – Hedgehog Oct 28 '14 at 14:13
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    This seems like a good reference for @bof's comment. – Pedro M. Oct 28 '14 at 17:58

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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.

5xum
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  • @bof Shouldn't this be a comment under the original question? – 5xum Oct 28 '14 at 13:53
  • @5xum It looks like we can conintue the process infinitely to constrcut countable familiy of $a_k$. First we have to carefully construct bounds for every $a_k^{(n)}$, so no bounds will intersect. Second we must have enough points. So for the first iteration we will have countable number of points, for the secound countable famility of countable number of points etc. Since countable union of countable sets is countable every iteration has countable number of points. And since we have countable number of steps total number of points will be countable. Can you verify my reasoning? – Hedgehog Oct 28 '14 at 14:06
  • If we iterate this process infinitely, in the sense that the set $A = {a^{(n_1, \ldots, n_j)}k}$ is constructed in such a way that all these numbers are distinct and $a^{(n_1, \ldots, n_j)}_k \to a^{(n_1, \ldots, n{j-1})}_{n_j}$, then every point of $A$ is a limit point, so there would be nothing to remove in the first iteration. – Pedro M. Oct 28 '14 at 18:05
  • @Pedro M. I have to think to understand why your last statement is true, so I can not answer you my ideas right now. – Hedgehog Oct 28 '14 at 18:12