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Let $A'$ denote the set of accumulation points of $A$. Find a subset $A$ of $\Bbb R^2$ such that $A, A', A'', A'''$ are all distinct.

I can find a set $A$ such that $A$ and $A'$ are distinct, but not one where $A,A',A'',A'''$ are all distinct.

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up vote 17 down vote accepted

Here’s a schematic of what you need to get distinct $A,A'$, and $A''$; it generalizes easily.

$$\begin{array}{rr} \bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\longrightarrow&\bullet\\ \uparrow&\uparrow&\uparrow&\uparrow&\uparrow&\uparrow\\ \color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\ \color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\ \color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\ \color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet\\ \color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet&\color{red}\bullet\color{red}\bullet\color{red}\bullet\color{red}\to\bullet \end{array}$$

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This picture is worth a thousand words (or a mess of formulas). – Andreas Blass May 24 '13 at 16:51
Cool! in LaTex.. – Henno Brandsma May 24 '13 at 16:55
@Asaf: Hey, I can only use what I know! I didn’t even know any $\LaTeX$ until I started posting here, not quite two years ago. – Brian M. Scott May 24 '13 at 18:09
Thank you for spending your time making the diagram! – John Michael May 24 '13 at 20:51
@John: You’re welcome! I like good visual aids. – Brian M. Scott May 24 '13 at 20:52

Hint: Work backwards. Start with some set, then add a sequence which converges to each point, then add a sequence which converges to each of the points in that set, and so on.

Do this carefully and you can make sure that $A$ is such set.

If you are familiar with the concept of ordinals, then you can also consider the ordinal $\omega^4$ embedded into the real numbers.

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I'm likely missing something: how is $A'$ different from the closure of $A$? – Martin Argerami May 24 '13 at 16:42
@MartinArgerami: if $A=\{1/2,1/4,1/8,\ldots\}$, $A'=\{0\}$. The closure of $A$ is the union of these. – Ross Millikan May 24 '13 at 16:47
@MartinArgerami The same thing confused me too, initially. It seems that (at least according to Wikipedia), $x$ is an accumulation point of $A$ if $x \in \overline{A\setminus\{x\}}$, i.e. there has to be a non-trivial sequence in $A$ which converges to $x$ – fgp May 24 '13 at 16:48
@fgp: That’s correct: every open nbhd of $x$ must contain some point of $A$ different from $x$ in order for $x$ to be an accumulation point of $A$. – Brian M. Scott May 24 '13 at 16:55
@fgp: I’ve seen them used as synonyms. I prefer to use cluster point only in the context of sequences (and nets), so that I can say that $1$ is a cluster point of $\sigma=\langle(-1)^n:n\in\Bbb N\rangle$ even though it’s not an accumulation (or limit) point of the range of $\sigma$. (German Wikipedia uses Häufungspunkt for both but carefully notes that * Folgenhäufungspunkt* and Mengenhäufungspunkt aren’t quite the same concept.) I do use accumulation point and limit point pretty much interchangeably, though the former has some special uses (e.g., complete accumulation point). – Brian M. Scott May 24 '13 at 17:14

What might be confusing here is that you won't, of course, get that $\overline{A} \neq \overline{\overline{A}}$ (where $\overline{A}$ denotes the closure of $A$). Thus, the accumulation points of $A'$ will also be accumulation points of $A$, but $A'$ may have fewer accumulation points than $A$ does.

In other words, you'll have that $A' \supset A'' \supset A''' \ldots$. This means that have to start with a set with "enough" accumulation points, otherwise you'll hit the empty set too early for $A',A'',A'''$ to be all different. The easiest way to avoid that is to work backwards, i.e. start with $A'''$, as Asaf suggested.

Let $$ A''' = \{(0,0)\} \text{.} $$ Then construct some $A'' \supset A'''$ such that $A''$ has accumulation point $(0,0)$, say $$ A'' = \{(\frac{1}{n},0) \,:\, n \in \mathbb{N}\} \cup \{(0,0)\} \text{.} $$ Now continue this idea to find $A'$. $A'$ will have to contain sequences which converge to $(\frac{1}{n},0))$ for every $n \in \mathbb{N}$. Finally, from $A'$, determine $A$.

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I have no idea what you are talking about. – John Michael May 24 '13 at 17:19
@JohnMichael I tried to explain the idea a bit more verbosely... – fgp May 24 '13 at 17:35
Sorry, there was an error and only half of your answer appeared! – John Michael May 24 '13 at 20:53
@JohnMichael No error, I expanded it after reading your comment, figuring I should explain things better... – fgp May 24 '13 at 21:02

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