Construct a bounded closed subset of $\mathbb{R}$ with exactly three limit points.

This might be a dumb question but why can't you just choose say $S=\{1,2,3\}$? It's bounded, and any sequence in $S$ that converges must eventually become constant. So any sequence must then converge to $1,2$ or $3$ which means $S$ is closed. Using the same reasoning, the limit points are $1,2$ and $3$.

I am using the sequential definition of "closed", and the following definition for the set of limit points of a given subset $S$ inside a metric space $M$: $$\lim S:=\{p \in M \mid \exists (x_n) \in S \text{ such that } x_n \to p\} .$$

  • $\begingroup$ I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = \{0\} \cup \{\frac{1}{n} : n \in \mathbb{N}\}$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = \{1\} \cup \{1 + \frac{1}{n}\}$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points. $\endgroup$ – Good Morning Captain Aug 15 '18 at 4:36
  • $\begingroup$ What you define is just the closure of $S$, not exactly the same. $\endgroup$ – Carsten S Aug 15 '18 at 8:51

With that definition, the set $S$ that you have actually works out, because every convergent sequence must be eventually constant, so must have as a limit one of the elements of $S$.

However, the definition of limit point in some other books is somewhat different :

$x \in S$ is a limit point if for all open sets $O$ with $x \in O$, there exists $q \neq x \in S$ such that $q \in O$. In other words, $(O \setminus \{x\}) \cap E \neq \emptyset$ for all open sets $O$ containing $x$.

The way of interpreting this definition, is that a limit point of $S$ is one such that at any arbitrary distance, one can find a point of $S$ closer than that distance which is different from $x$. In other words, there is a sequence, none of whose terms are $x$, which converges to $x$.

For example,Rudin has this definition.

Now, with this definition $S$ does not work, because any eventually constant sequence cannot be used to show that some point is a limit point. We have to, instead, be cleverer.

Indeed, the following set works out now : $\{\frac 12,\frac 23,\frac 34,\frac45,...,1\} \cup \{1\frac 12,1\frac 34 , 1 \frac 45,...,2\} \cup \{2\frac 12,2\frac 23,2\frac 34,...,3\}$.This set has exactly three limit points, according to the new definition.

You must not be too worried that your definitions are not tallying with other ones. Proceed with your definitions, and soon enough you will comfortable in switching while having conversations with others who have differing definitions.


For any $m$ distinct reals (or any domain where the following makes sense) $(a_k)_{k=1}^m $, the $m$ sequences $S_k = (a_k+\frac1{n})_{n=1}^{\infty} $ for $k=1$ to $m$ are bounded and $S_k$ converges to $a_k$.

You can interlace these to get a single sequence with $m$ subsets that each converge to one of the $a_k$: $T_n = (a_{1+(n-1 \bmod m)} +\frac1{m\lfloor (n+m-1)/m \rfloor} $ for $n=1$ to $\infty$.


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