Construct a set of the real line such that nth derived set is empty.

Question

I am self studying from the book Elementary Real and Complex Analysis but need some help with the following exercise.

Let $A'$ denote the set of all limit points of a given subset $A$ of a metric space $M$ and let $A^{(n)}=(A^{(n-1)})'$. Given any $n$, construct a set $A$ on the real line such that $A^{(n)}$ is nonempty while $A^{(n+1)}$ is empty.

First, I noticed that for $A= \{\frac{1}{M} \mid M \in \mathbb{N}\}$, then $A'=\{0\}$ and $A'' = \emptyset$. Therefore, I proposed the set $A=\{\frac{1}{m_1} + \cdots + \frac{1}{m_n} \mid m_i \in \mathbb{N} \}$ in order to reach the simple case at some step. While it is clear that any sum $\frac{1}{m_1} + \cdots + \frac{1}{m_{n-1}}$ is a limit point of $A$, I cannot prove that there are not other limit points.

I also failed to prove this by induction.

Answer 1

You do want to start with a simple sequence. Then make each point of the sequence the limit of another simple sequence. Then make each of the newly added points the limit of yet another simple sequence. Keep going. Here’s one way to do it all at once; I’ll leave it to you to verify the details.

For $n\ge 1$ let $\Sigma_n$ be the set of $n$-tuples of positive integers. For each $\langle k_1,\ldots,k_n\rangle\in\Sigma_n$ let

$$x_{\langle k_1,\ldots,k_n\rangle}=2^{-k_1}+2^{-(k_1+k_2)}+\ldots+2^{-(k_1+k_2+\ldots+k_n)}\;.$$

Let $L_n=\{x_{\langle k_1,\ldots,k_n\rangle}:\langle k_1,\ldots,k_n\rangle\in\Sigma_n\}$.

Finally, for $n\ge 0$ let

$$A_n=\{0\}\cup\bigcup_{k=1}^{n+1}L_k\;.$$

Show that $A_0'=\{0\}$, and $A_n'=A_{n-1}$ for $n\ge 1$.

I suggest drawing pictures of the first two or three of these sets to get a better idea of what’s going on here. You will find it useful to show that $x_{\langle k_1,\ldots,k_n,k\rangle}<x_{\langle k_1,\ldots,k_n-1\rangle}$ for all $k\ge 1$ and $k_n\ge 2$.