13

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.

Brian M. Scott
  • 616,228
John Michael
  • 493

3 Answers3

21

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}$$

Brian M. Scott
  • 616,228
10

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.

Asaf Karagila
  • 393,674
  • I'm likely missing something: how is $A'$ different from the closure of $A$? – Martin Argerami May 24 '13 at 16:42
  • 1
    @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
  • @BrianM.Scott The interesting thing is, I'm pretty sure that this isn't the definition I learned years ago. Then again, my native language is german, so maybe there's a subtle difference between "Häufungspunkt" and "accumulation point", even though the former is pretty much the direct translation of the latter. Out of curiosity - do you distinguish between "accumulation point", "limit point" and "cluster point", or are these all synonyms? 'cause they all seems to translate to "Häufungspunkt".. – fgp May 24 '13 at 17:02
  • 1
    @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
2

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 $A$.

fgp
  • 21,050