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

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

Please help to decide whether this solution is correct:

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

An illustration of some points in the subset:

enter image description here

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.

  • $\begingroup$ Yes, very nice ! $\endgroup$
    – Asinomás
    Dec 14, 2016 at 17:43
  • $\begingroup$ 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 $\eta$.) $\endgroup$ Dec 14, 2016 at 17:57

1 Answer 1


We can generalize:

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 $a_2,a_3,\dots$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.