Can a bounded set of isolated points contain infinitely limit points in a complete metric space?

Question

This is not the problem I am trying to solve, but if this result were true, then the problem would be easier to solve.

Let ($X$, $d$) be a complete and totally bounded metric space and let $B$ $\subseteq$ $X$ be a bounded set of isolated points. If it happens that B is an infinite set, then $B$ must contain limit point(s) by boundedness of $B$. Please note that we cannot assume that $B$ is compact.

So define $C$ = {$y$ $\in$ X | $y$ is a limit point of $B$}. It is clear from the definition of isolated point that there are no points of $B$ in $C$.

My question is, Is $C$ necessarily a finite set?

As I mentioned in the title of the post, I would like the answer to be, yes, but I am not convinced that this may be true.

So if I start out by assuming that $C$ contains infinitely many points in $X$, then $C$ itself must have limit point(s) otherwise $C$ would be unbounded, which would then imply that $B$ is unbounded. So let $y_0$ be some limit point of C. It is clear that $y_0$ $\in$ $C$ since for any neighborhood $N$ of $y_0$, there exist points of C, not equal to $y_0$ that are in $N$. If $a$ $\in$ $C$ is such a point, there are, of course points in $B$ arbitrarily close $a$. Hence $y_0$ is a limit point of B, so by definition of $C$, $y_0$ $\in$ $C$. From this it follows that $C$ contains its limit points. If we are to prove that $C$ must be finite, one way would be to somehow show that there must exist a point $b$ $\in$ $B$ such that $b$ is a limit point of $B$, i.e. $b$ $\in$ $C$. Or equivalently, can we find some subsequence in $B$ that converges to a point in B. This would be a contradiction.

If it is true that $C$ can be infinite, must the set of limit points of that set be finite, etc.?

Answer 1

It's certainly possible to have infinitely many limit points of $B$. Consider a set of convergent sequences, all disjoint with distinct limits (and $B$ is the set of sequence points without those limits, and the limits of the sequences are the limit points of $B$).

Answer 2

It is possible for $C$ to be not just infinite but uncountable. For instance, let $X=[0,1]^2\subset\mathbb{R}^2$, and let $$B=\{(1/n,m/n):m,n\in\mathbb{Z}_+,0<m<n\}.$$ It is easy to see that $B$ is bounded and discrete. However, every point of the form $(0,t)$ for $t\in[0,1]$ is a limit point of $B$, since you can choose arbitrary large $n$ and approximate $t$ by a rational number of the form $\frac{m}{n}$. So in this case $C=\{0\}\times[0,1]$ is uncountable (and moreover, every point of $C$ is a limit point of $C$).

Answer 3

Let $X=[-1,1]^2\subset \Bbb R^2$ and $B=\{\,(1/n,1/m)\mid n,m\in\Bbb N\,\}$. Then $C$ contains all $(1/n,0)$, among others.

Answer 4

Take any countable successor ordinal $\alpha$. Then $\alpha$ with the order topology is compact and second-countable, so it is completely metrizable (and also bounded in any metric you choose to put on it, by compactness).

If you look at all successor ordinals in $\alpha$, then their limit points will be exactly the limit ordinals in $\alpha$, i.e. multiples of $\omega$. If you look at the limit points of that (or even just the isolated points there), the limit points will be the multiples of $\omega^2$, and so on. In this way (by taking sufficiently large $\alpha$, you can get an arbitrarily long (countable well-ordered) sequence of discrete sets, each consisting of (some) limit points of the previous one.

For example, if $\alpha$ is $\omega^2+1$ and $B$ is the set of successor ordinals, then $C$ is infinite and its only accumulation point is $\omega^2$, so $C\setminus \{\omega^2\}$ is discrete.