Q: Derivative of the derivative of a set Suppose that we have a non-empty set called $X\subseteq \Bbb R$. The derivative $DX\neq \emptyset$ (I not want to consider particular cases - see the comments -) of a set is constituted by the accumulation points of $X$.
But if I consider $D(D(X))$, i.e. the derivative of the derivative of $X$, surely it's still a derivative set.

Is it possible it could be $D(D(X))\subseteq DX$? What happens if is there a sequence of derivative sets? In other words, is it possible that
$$D(D(D(\cdots(X))))=\emptyset$$
if $X\neq \emptyset$ or is there always an empty-set $Y=D(D(D(\cdots(X))))\neq\emptyset$?


A: As mentioned in the comment, googling Cantor-Bendixson rank may help you, and if you know something about ordinal numbers you may find it easier to understand.
I don't understand your question very well. $D(D(X))\subseteq D(X)$ always holds for $X\subseteq\mathbb{R}$ because $D(X)$ is a closed set. I guess you may be asking whether for any $n$ there exists some $X$ such that $D^{(n)}(X)\neq\emptyset$, where $D^{(n)}(X)$ means taking derivative repeatedly for $n$ times. This is indeed possible. For some complicated $X$ it may even happen that $D^{(n)}(X)\neq\emptyset$ for every $n$; furthermore the intersection of all these things, namely $\bigcap_{n=1}^{\infty}D^{(n)}(X)$, could be non-empty; call this intersection $D^{(\omega)}(X)$ and we can keep iterating, obtaining $D^{(\omega+1)}(X)$, $D^{(\omega+2)}(X)$, $D^{(\omega+\omega)}(X)$, etc.
To see why is that possible, let us first consider the simplest case when $D(X)$ contains exactly one point. An example is $X=\{-1,-\frac{1}{2},-\frac{1}{4},...,-\frac{1}{2^m},...\}\cup\{0\}$. All those negative points are isolated, though they become closer and closer to each other until they "cluster" at $0$. Now $D(X)$ means to throw away all isolated points, so we have the point $0$ left, i.e., $D(X)=\{0\}.$ If we take the derivative of $D(X)$, since $0$ is clearly isolated in $\{0\}$, we have nothing left, $D(D(X))=\emptyset$.
Now to get some $Y$ such that $D(D(Y))\neq\emptyset$, we "place a copy of $X$ at every isolated point of $X$" ($X$ is same as before). It is better to imagine the picture; if written down $Y$ could be something like $X\cup\{-\frac{1}{2^m}-\frac{1}{2^{m+k}}\mid m,k\geq 1\}$. Every point in the original $X$, say $-\frac{1}{2^m}$, becomes non-isolated because it is approached by the sequence $(-\frac{1}{2^{m+k}})_{k\geq 1}$ from the left ($0$ of course remains non-isolated in $Y$), while all these new points $-\frac{1}{2^{m+k}}$ are isolated. Therefore $D(Y)=X$, and by previous paragraph we see $D(D(Y))=D(X)=\{0\}$, and $D(D(D(Y)))=\emptyset$. Using this method we can obtain set $Z$ such that $D^{(n)}(Z)$ vanishes exactly at any desired stage $n$.
To get an $X$ such that $D^{(n)}(X)$ is non-empty for every $n$, we just take the union of previously constructed $X_n$ such that $D^{(n)}(X_n)$ is non-empty, that is $X=\bigcup_{n=1}^{\infty}X_n$ ,and notice that since $X\supseteq X_{n}$ for every $n$, and since $A\supseteq B$ implies $D(A)\supseteq D(B)$, we have $D^{(n)}(X)$ non-empty for every $n$.
In essence, what we are doing is just mapping countable ordinal numbers into the real line. The order type of the $X$ defined above is called $\omega+1$, because there are $\omega$ or infinitely many points (those $-\frac{1}{2^{m}}$'s) followed by one point (which is $0$). The order type of $Y$ is $\omega$-many $X$'s followed by a point, and therefore is $\omega\times\omega+1$, also denoted as $\omega^{2}+1$, etc. By the so called  transfinite induction we can map any countable ordinal $\alpha$ into the real line, and thus obtain $X$ such that $D^{(\alpha)}(X)$ vanishes exactly at any desired stage $\alpha$.
