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I am working on the following problem from Bergman's supplement to Rudin PMA:

Given a set $E$, let $L(E)$ denote its limit points, $L^n(E) = L(\dots(L(E)))$, and $L^0(E) = E$. Construct for every $n \in \mathbb{Z}^+$ a subset $E \subseteq \mathbb{R}$ such that $L^{n - 1}(E) \ne \emptyset$, but $L^n(E) = \emptyset$.

Apparently the construction should be given recursively, and so far my idea is to fill the gaps between each element in the previous iteration with a convergent sequence, which should get eliminated upon taking the limit points.

Let $E_1 = \{0\}$, $E_2 = \{\frac{1}{n} : n \in \mathbb{Z}^+\}$. Having constructed countable $E_3, \dots, E_{n - 1}$, we arrange each $E_m$ into a decreasing sequence $(e^m_i)$, and let $$E_n = \bigcup_{i = 1}^\infty \{e^{n - 1}_i + \frac{1}{n} : n \in \mathbb{Z}^+ \land \frac{1}{n} < e^{n - 1}_{i + 1} - e^{n - 1}_i\}, $$ which is countable as a countable union of countable sets. We claim that $L(E_n) = E_{n - 1}$, in which case we would be done, as $L(E_1) = \emptyset$.

The main problem is that I don't know how to make "arranging into a decreasing sequence" a rigorous argument, especially since we are restricted to the canonical ordering.

n1lp0tence
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1 Answers1

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You are on the right track, however I would slightly tweak that construction. Let $E_0=\{0\}$ and then given $E_n$ take the set $d(E_n)$ of all isolated points (in $E_n$). For each $x\in d(E_n)$ consider $a_n=x+1/(n+m)$ where $m$ is big enough so that $(x-1/m,x+1/m)\cap E_n=\{x\}$. Such $m$ exists because $x$ is isolated. Finally define $E_{n+1}$ as union of $E_n$ with all $a_n$ sequences over all isolated points of $E_n$.

The main observation of this construction is that at each step we turn all isolated points into non-isolated by adding isolated points only, i.e $d(E_{n+1})=E_{n+1}\backslash E_n$. Thus $L(E_{n+1})=E_n$ should be straightforward.

freakish
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