Let $A$ be a set in the real line $\Bbb R$, and $A'$ the derived set of $A$, and $A''$ the derived set of $A'$, and so on. Is it possible to get an infinitively many distinct subsets of $\Bbb R$?
And in a finite dimensional normed vector space ?
Thank you very much.
Yes, to both. And this is possible in any space that includes a copy of the rationals (we can already do it in that). This involves a bit of set theory (and it is the reason ordinals were discovered/invented by Cantor originally!):
If we start with a set $A$, we define $A^{(0)} = A$, and $A^{(\alpha+1)} = (A^{(\alpha)})'$ for successor ordinals, and $A^{(\beta)} = \cap \{A^{(\alpha)}: \alpha < \beta\}$ for limit ordinals $\beta$. Then for some ordinal $\gamma$, we have $A^{(\gamma)} = A^{(\gamma+1)}$ and we stabilise. The minimal $\gamma$ for which this happens is called the scattering height of the set $A$. For subsets of separable metrisable spaces (like $\mathbb{R}$, or finite powers of it), this ordinal is countable, i.e. $\gamma < \omega_1$. But we can get any countable successor ordinal we like as the scattering height, so in particular we can have infinitely many different successive derived sets.
In fact, ordinal numbers themselves (seen as ordered topological space) can achieve this already, and all countable ordinal numbers embed topologically into $\mathbb{Q}$, hence my first remarks.
Here is an explicit example. For each $n\in\Bbb Z^+$ let
$$S_n=\{\langle k_1,\ldots,k_n\rangle\in\Bbb (Z^+)^n:k_1<\ldots<k_n\}\;,$$
the set of strictly increasing $n$-tuples of positive integers. Let $S=\bigcup_{n\ge 1}S_n$, and define
$$\varphi:S\to(0,1):\langle k_1,\ldots,k_n\rangle\mapsto 2^{-k_1}+\ldots+2^{-k_n}\;.$$
Note that $\varphi$ is injective, and for any $s=\langle k_1,\ldots,k_n\rangle\in S$, the sequence
$$\big\langle\varphi(\langle k_1,\ldots,k_n,\ell\rangle):\ell\ge 1\big\rangle$$
is a strictly decreasing sequence of real numbers converging to $\varphi(s)$. Finally, the sequence $\langle\varphi(\ell):\ell\in\Bbb Z^+\rangle$ converges to $0$.
For $n\in\Bbb Z^+$ let $A_n=\{0\}\cup\bigcup_{k=1}^n\varphi[S_k]$. $A_1$ is a simple sequence converging to $0$ together with its limit $0$. $A_2$ is obtained from $A_1$ by adding a simple sequence converging to each isolated point of $A_1$, and in general $A_{n+1}$ is obtained from $A_n$ by adding a simple sequence converging to each isolated point of $A_n$. (It helps to sketch the first three or so of these sets.) Thus, the derived set of $A_1$ is $\{0\}$, and for $n>1$ the derived set of $A_n$ is $A_{n-1}$.
Let $$X_n=A_n+(n-1)=\{x+n-1:x\in A_n\}\;,$$ and let $X=\bigcup_{n\ge 1}X_n$; then
$$\begin{align*} X'&=\{0\}\cup(X+1)\;,\\ X''&=\{1\}\cup(X+2)\;, \end{align*}$$
and in general for $n\ge 1$ we have
$$X^{(n)}=\{n-1\}\cup(X+n)\;.$$
Clearly these sets are distinct.
Yes. To understand this better, though, you need to know something about ordinals.
If $A$ is just one convergent sequence (with its limit), then $A''$ is empty. But if $A$ is a collection of sequences, whose limit points make a convergent sequence (e.g. for each $\frac1n$ attach a convergent sequence) then $A''$ is not empty anymore.
You can iterate this process, until you reach a set which has infinitely many distinct iterated derived sets. And then you can take the intersection of all those sets, and continue.
But in $\Bbb R$ you cannot continue this process uncountably many times. At some point you will have to stabilize.