# Compact subset of $\mathbb{R}^1$ with countable limit points

Question 13 of chapter 2 in "Principles of mathematical analysis" by Rudin:

Construct a compact subset of real numbers which has countable limit points.

$A=\{ 1/2^n \} \bigcup \{1/2^n + 1/2^m\} \bigcup \{0\}$ where $n=1, 2, \cdots$ and $m$ is an integer such that $m>n$

An illustration of some points in the subset:

This set is bounded as $\forall x \in A \implies 0 \le x<1$.

Notice that the interval $(1/2^n+1/2^{(n+1)}, 1/2^{(n-1)})$ contains no point in set $A$. Since this is an open interval each point in this set has a neighbourhood which does not contain anypoint of A. Therefore this interval contains no limit points.

Further each interval of the form $((1/2^n + 1/2^{m+1}),(1/2^n + 1/2^m))$ does not contain any point of A so being open intervals they have no limit points.

Thus no limit point of $A$ lies outside $A$. Therefore $A$ is closed.

Being closed and bounded $A$ is compact.

The points $\{ 1/2^n \} \bigcup \{0\}$ where $n=1, 2, \cdots$ are limit points as each neighbourhood contains point of $A$.

On other hand points $\{1/2^n + 1/2^m\}$ where $m>n$ are not limit points as they have intervals on both side which do not contain any point of $A$.

The set $\{ 1/2^n \} \bigcup \{0\}$ is countable (can be mapple to tupple (n,m which is countable)) therefore $A$ is a compact set having limit points which form a countable set.

• Yes, very nice ! – Jorge Fernández Hidalgo Dec 14 '16 at 17:43
• Some terminology you may or may not care about — the order type of this set is called $(\omega^2)^*$. $~\omega$ is the order type of the positive integers; $\omega^*$ is the reverse of this (and the order type of the negative integers). $~\omega^2$ is $\omega+\omega+\omega+\dotsb$, and its reverse, $(\omega^2)^*$, is the order type of your set. (The order type of the rationals is sometimes called $\eta$.) – Akiva Weinberger Dec 14 '16 at 17:57

Take an increasing sequence of positive reals $a_1,a_2,\dots$ converging to $l$.
For each $i>1$ we let $(b^i)$ be an increasing sequence of reals converging to $a_i$ such that $b^i_j>a_{i-1}$.
Your same arguments prove $(a) \cup ( \bigcup\limits_{i=1}^\infty (b_i))$ is closed and compact and has limit points $a_2,a_3,\dots$