let denoted $A'$ Derived set,and define $$A'=A^{(1)},(A^{(1)})'=A^{(2)},\cdots,(A^{(n)})'=A^{(n+1)}$$
Question:
Take example the set $A$,such $A^{(k)},k=1,2,\cdots,n+1$ are different each other
I have consider sometimes,and try take some example and at last,I found my example not such condition. Thank you for help
PS: this problem is from real analysis book problem excise page 201(by Lin sen XU).
The standard approach uses ordinals. First, check that given any countable ordinal (with the order topology) there is a set of reals homeomorphic to it (in fact, you can pick the set to consist entirely of rationals, and you can ensure that, as an ordered set, it is also isomorphic to the ordinal). This is an easy (transfinite) inductive argument, using that all intervals $(p,q)\cap\mathbb Q$ (with $p<q$ rationals) are homeomorphic.
The point is that it is easy to compute the (iterated) Cantor-Bendixson derivatives of ordinals. For instance, if $\alpha=\omega^2+1$, then $\alpha'$ is homeomorphic to $\omega+1$ and $\alpha''$ is a singleton. If $\alpha=\omega^\omega+1$ (ordinal exponentiation), then $\alpha,\alpha',\alpha'',\dots,\alpha^{(n)},\alpha^{(\omega)}$ are all different, with $\alpha^{(\omega)}$ being a singleton.
[For limit ordinals $\alpha$, we define $A^{(\alpha)}$ as $\bigcap_{\beta<\alpha}A^{(\beta)}$, while for successor ordinals we just iterate, as in the finite case. In general, if $\alpha=\omega^\beta+1$ (ordinal exponentiation) with $1\le \beta<\omega_1$, then $\alpha^{(beta)}$ is a singleton, and all sets in the (long) sequence $\alpha,\alpha',\dots,\alpha^{(\beta)}$ are distinct.]
Being explicit, all you are doing is this: Fix a increasing, converging sequence of reals, $a_0<a_1<\dots<a=\lim_n a_n$. Now fix, in $(a_0,a_1)$, a strictly increasing sequence, converging to $a_1$. In $(a_1,a_2)$, do the same, fix a strictly increasing sequence converging to $a_2$, say, $b_0<b_1<\dots$ but, in addition, in each $(b_i,b_{i+1})$, fix a strictly increasing sequence, converging to $b_{i+1}$. Do the same, but iterate the process one more time in $(a_2,a_3)$, and then yet one more time in $(a_3,a_4)$, etc. The set $A$ consisting of all the points you used in this construction is as desired.
In a sense, the use of ordinals is inevitable. For instance, Baumgartner proved that if $X$ is a countable Hausdorff space with a countable basis, $1\le \alpha<\omega_1$, and $X^{(\alpha)}\ne\emptyset$, then $X$ contains a subspace homeomorphic to $\omega^\alpha+1$. (See Theorem 2.1 in J. Baumgartner, Partition relations for countable topological spaces, Journal of Combinatorial Theory, Series A, 43, (1986), 178-195. In fact, Baumgartner showed a technical strengthening of the statement I made, namely, that this holds even if $X$ does not admit a countable basis, as long as its weight is less than the cardinal invariant $\mathfrak p$.)