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I want to solve the following exercise 1.3.C. from page 27 of R. Engelking: General Topology.

For every positive integer $n$ the $n$-th derived set $A^{(n)}$ of a subset $A$ of a topological space $X$ is defined inductively by the formulas: $$ A^{(1)} = A^d \quad \textrm{and} \quad A^{(n)} = (A^{(n-1)})^d. $$ (a) Give an example of a set of real numbers that has three consecutive derived sets distinct from each other.

(b) Give an example of a set of real numbers that has infinitely many derived sets distinct from each other.

The definition is $$ A^d := \{ x \in X \mid x \in \overline{A \setminus \{ x \}} \}. $$ I am confused, cause the set $A^d$ ist closed, i.e. $\overline{A^d} = A^d$. And closed means it already contains all its limit points, so how can then $(A^d)^d$ be different from $A^d$?

amWhy
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StefanH
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    The picture in this answer to a question very similar to (a) should help in visualizing what’s going on, if help is needed. For (b), you can take a discrete union of spaces with larger and larger finite Cantor-Bendixson ranks. – Brian M. Scott Jul 02 '13 at 19:26

2 Answers2

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To address your confusion about the closedness of derived sets: Note that $A^d$ being closed simply means that $(A^d)^d$ is a subset of $A^d$. (Actually $A^d\subseteq A$ is equivalent to $A$ being closed for any set $A$.) So this says that a limit point of $A^d$ is also a limit point of $A$, but a limit point of $A$ need not be a limit point of $A^d$, or in other words, $(A^d)^d$ can be a proper subset of $A^d$.

Now regarding your question about a subset with infinitely many distinct derived sets, try the following.

Let $B=\{0\}\cup\{\frac1n\mid n\in\mathbb N\}$.

For $n\ge2$ define $B_n^1=\{\frac1n\}\cup\{\frac1n+\frac1k(\frac1{n-1}-\frac1n)\mid k\ge2\}$ and let $b_n^1$ be the point $\frac1n+\frac12(\frac1{n-1}-\frac1n)$, the right most point of this sequence.

Define $B_n^{l+1}$ for $1\le l<n$ recursively as $B_n^{l+1}=\{b_n^l+\frac1k(\frac1{n-1}-b_n^l)\mid k\ge2\}$
and $b^{l+1}_n$ as the point $\frac12(\frac1{n-1}-b_n^l)$

Now you can take the union $$B\ \cup\ \bigcup\limits_{1 \leq l < n \\ n \in \mathbb{N}}{B_n^l}$$

If I'm not mistaken this should yield a set with infinitely many distinct derived sets.

You should draw a picture for the first few $l$ to get an idea how this set will look like.

pabhp
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Stefan Hamcke
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A start: Consider first the set $P$ consisting of $0$ and all $\frac{1}{n}$, where $n$ ranges over the integers $\ge 2$.

The derived set of $P$ is $\{0\}$.

Now for every positive element $\frac{1}{n}$ of $P$, add a sequence that approaches $\frac{1}{n}$ from the right. For definiteness, the sequence for $\frac{1}{n}$ consists of the numbers $\frac{1}{n}+\frac{1}{k}\left(\frac{1}{n-1}-\frac{1}{n}\right)$, where $k$ ranges over the integers $\ge 2$.

Let $Q$ be $P$ together with these added points. The derived set of $Q$ is $P$, and the derived set of $P$ is $\{0\}$. This example should answer your $(A^d)^d$ question, and give a start towards the construction of examples that the problem asks for.

Note for example that $Q^d$, $(Q^d)^d$, and $((Q^d)^d)^d$ are distinct.

André Nicolas
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  • yes, I understand with your construction I can build for every $k$ a sequence such that the first $k$ derived sets are distinct. But regarding (b) I have somewhere read about the Cantor-Bendixson rank which says that every sequence of derived sets gets stationary, so how does this relate to the exersice asking for an infinite progression that never gets stationary, i.e. where $A^{(n)} = A^{(n+1)}$ for no $n$? – StefanH Jul 02 '13 at 16:29
  • You can iterate the construction of my example forever. That's enough to deal with your full problem. Now it so happens that we can continue the construction into the transfinite. So after doing it "forever" you can do it again to the resulting set. This is of enormous historical importance, since this is what led Cantor to discover the infinite ordinals. (And, contrary to popular belief, infinite ordinals came first, not cardinals.) – André Nicolas Jul 02 '13 at 16:35
  • Wouldn't this give a set where every point turned out to be a limit point sinc every point is approached by a sequence from the right. – Stefan Hamcke Jul 02 '13 at 16:37
  • You are right, I was sloppy in my comment. Do it this way. Above $0$, put a set like my (renamed, so as not to interfere with your notation) $P$. Above $1$, put a set like my $Q$. Above $3$ put a set whose derived set is like $Q$. And so on. Then as we derive $n$ times, the stuff up to $n$ dies. – André Nicolas Jul 02 '13 at 16:45