Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 16 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}$$

share|improve this answer
5  
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
1  
@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
1  
Thank you for spending your time making the diagram! –  John Michael May 24 '13 at 20:51
1  
@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.

share|improve this answer
    
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
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

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 $A$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.