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I have a doubt on the limit set.

If we call limit set S' the set of all limit points of a given set S, is it by any means true that, regardless of the initial set S, we have S''' = S'' ? In other words, is it true that if we iterate the operation more than twice we'll always keep having the same set we got after two iterations? Or can we go on indefinitely and still possibly get a different set at every iteration?

If S''' = S'', could you please provide me with a proof? I've tried to google it but I couldn't find anything.

Thank you.

keine_ahnung
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1 Answers1

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I also tried googling and google AI claims that "the limit points of the limit points of a set of real numbers are simply the original limit points of the set". I don't think google AI is correct based on the following counterexample.
For positive integer $k$, define the following set on the real line with respect to the standard topology: $$A_k= \{ \frac{1}{n_1}+\frac{1}{2n_1n_2}+\ldots +\frac{1}{2^{k-1}\Pi_{j=1}^k n_j }: n_1<n_2<\ldots <n_k \in \mathbb{Z}^+ \} .$$ The only limit point of $A_1$ is 0. The limit point set from $A_2$ is $A_1\cup \{ 0\}$. In general, the limit point set of $A_k$ is $$\Big( \bigcup_{i=1}^{k-1}A_{i}\Big) \cup \{0\} .$$
Define the set $B_k$ as $$B_k = \Big( \bigcup_{i=1}^{k} A_i \Big) \cup \{ 0\}.$$ Therefore, if we iteratively take limit points $\ell$ times of the set $B_k$, we get $$B_k^{(\ell)} = B_{k-\ell} .$$ This generates a different set for $\ell=0,1,\ldots ,k-1.$

The Other Terry
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