# Construct a compact set of real numbers whose limit points form a countable set.

I searched and found out that the below is a compact set of real numbers whose limit points form a countable set. I know the set in real number is compact if and only if it is bounded and closed. It's obvious it is bounded since $\,d(1/4, q) < 1\,$ for all $\,q \in E.$

However, I'm not sure how this is closed.

Is there any simpler set that satisfies the above condition?

Thank you!

$$E = \left\{\frac 1{2^m}\left(1 - \frac 1n\right) \mid m,n \in \mathbb N\right\}.$$

• Note that $1/2$ is a limit point of your set but is not in your set. – André Nicolas Jul 16 '13 at 0:17
• oops. You are right. d(1/4, q) < 1 for all q! – InfimumMaximum Jul 16 '13 at 0:51
• You mean "countably infinite", right? – Stefan Smith Jul 16 '13 at 1:39

What about $$A=\left\{\frac1n+\frac1 m:m,n\in\Bbb N\right\}\cup\{0\}\text{ ? }$$

One can see the $A'$ is $$\left\{\frac 1 n:n\in\Bbb N\right\}\cup \{0\}$$ Thus, let $E=A\cup A'=\bar A$ which is closed, and bounded.

• @DavidMitra Thank you, yes. – Pedro Tamaroff Jul 16 '13 at 0:35
• The set $\{\frac{1}{n}:n\geq 1\}\cup\{0\}$ is definitely in $A'$. The fact that these are the only points in $A'$ might need some effort to prove. – Jack Oct 3 '13 at 20:40
• Any ideas on how to prove those are all the limit points? – Marco Flores Sep 15 '14 at 1:46

The idea you used is good. I will do the same thing more slowly, building the set step by step in order to retain control over the geometry.

The backbone of the set is, like yours, the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$. This sequence has limit $0$, so we throw in the number $0$.

Now near $\frac{1}{n}$ for every $n$, we produce a sequence that has limit $\frac{1}{n}$, and that does not mix in unpleasant ways with other sequences. Note that the distance from $\frac{1}{n}$ to $\frac{1}{n+1}$ is $\ge \frac{1}{2n^2}$.

So for every $n$, throw in the numbers $\frac{1}{n}+\frac{1}{(2n^2)(2^m)}$, where $m\ge 1$.

Another way: Start with the set $A$ of numbers $n+\frac{1}{2^m}$, where $n$ and $m$ range over the positive integers. This set does the job beautifully. Only one minor flaw: the set is not compact. Let $B$ be the set of all reciprocals of numbers in $A$, together with $0$. This does the job.

The limit points are $\{\frac{1}{2^m}\mid m\in \mathbb{N}\}$. These are contained in the set (to get $\frac{1}{2^k}$ (for $k>1$), take $m=k-1$, $n=2$).

We can tell there are no other limit points, since the closest points to $\frac{1}{2^k}(1-\frac{1}{l})$ (for $l>2$) are $\frac{1}{2^k}(1-\frac{1}{l+1})$ and $\frac{1}{2^k}(1-\frac{1}{l-1})$, so we can isolate them in a neighborhood of radius $\frac{1}{2^k}(1 - \frac{1}{2(l+1)})$.

Edit: As Andre has pointed out, $1/2$ is not in the set, so the problem does not work as stated.

• Very good, @Eric, but what about $k=1$? Oh yeah, and $0$? – Ted Shifrin Jul 16 '13 at 0:16

What about the trivial unitary set as in $\{ 0 \}$?

• That is correct, and I believe you should add that the limit point set of the wished set must be non-empty. – Marra Jul 16 '13 at 0:28
• I would think "countable" means "countably infinite", in the OP. This should also be mentioned in the OP. – David Mitra Jul 16 '13 at 0:30
• That would make more sense. – Marra Jul 16 '13 at 0:31

For each positive integer m, put Aₘ = {2⁻ᵐ} ∪ {2⁻ᵐ+2⁻ⁿ , m,n∈ N⁺}. Am has only one limit point in 2⁻ᵐ. Put A = {0} ∪ (∪₁infinity Aₘ) For any fixed positive integer m, since 2⁻ᵐ⁻¹+3⁻ⁿ<2⁻ᵐ it is clear that A has no other limit points apart from the limit points S={0,1/2,1/4,...................} Apparently, S is countable. It remains to prove that A is compact. Since A ⊂ [0, 1], A is bounded. Also, S ⊂ A implies that A is closed