# To prove: if X' is not compact, then there is a countably infinite discrete set in X.

In a research article I found:

In a metrizable topological space $$(X,\tau)$$, if set of limit points $$X'$$ is not compact, then $$X'$$ contains a countably infinite discrete set.

I don't understand how it is happening.

If $$X'$$ is not compact, then there is a net $$\{x_\alpha\}_{\alpha\in J}$$ which has no convergent subnet. So, the net $$x_\alpha$$ has no limit point. Hence, the set $$S=\{x_\alpha:\alpha\in J\}$$ is a discrete space in the subspace topology $$\tau_S=\{S\cap O:O\in \tau\}$$, which is equivalent to each singleton set $$\{x_\alpha\}$$, $$\alpha\in J$$, is in $$\tau_S$$''. But from here how to prove that each singleton set $$\{x_\alpha\}$$, $$\alpha\in J$$, is in $$\tau$$.

A set is called discrete, if it consists only isolated points.

• You can use > arrows like this to format a block quote. May 18, 2021 at 5:13
• Sequences are adequate in metric spaces. Why do you use nets? May 18, 2021 at 5:27
• A net without a convergent subnet doesn't always give a discrete subset of $X$, so the general net idea is flawed. BTW every infinite metric space has a countably infinite discrete subset (it's a closed discrete subset that contradicts compactness). May 19, 2021 at 6:39

SKETCH: The space $$X$$ is largely irrelevant: you’re really just trying to show that if a space $$X$$ (the $$X'$$ of your question) is metrizable and not compact, then it contains a countably infinite discrete set. Let $$d$$ be a compatible metric on $$X$$. Since $$X$$ is not compact, either the metric space $$\langle X,d\rangle$$ is not complete, or it is not totally bounded.

If it is not complete, there is a $$d$$-Cauchy sequence $$\langle x_n:n\in\Bbb N\rangle$$ in $$X$$ that does not converge to any point of $$X$$. Show that $$\{x_n:n\in\Bbb N\}$$ is an infinite closed discrete subset of $$X$$.

If $$\langle X,d\rangle$$ is not totally bounded, there is an $$\epsilon>0$$ such that if $$F\subseteq X$$ is finite, then $$\bigcup_{x\in F}B(x,\epsilon)\ne X$$. Recursively choose points $$x_n$$ for $$n\in\Bbb N$$ as follows. Let $$x_0\in X$$. $$B(x_0,\epsilon)\ne X$$, so there is an $$x_1\in X\setminus B(x_0,\epsilon)$$. If $$n\in\Bbb N$$, and we’ve chosen $$x_k\in X$$ for each $$k\le n$$, $$\bigcup_{k\le n}B(x_k,\epsilon)\ne X$$, and we can keep the recursion going by choosing $$x_{n+1}\in X\setminus\bigcup_{k\le n}B(x_k,\epsilon)$$. Show that $$\{x_n:n\in\Bbb N\}$$ is an infinite discrete subset of $$X$$.

• $X= (-\infty, 0)$ is neither complete nor totally bounded, and it does not contain any isolated point.
– GGI
May 18, 2021 at 6:10
• @GripGrip: True, but in the present problem we don’t know what $X$ is: we just know that we have a Cauchy sequence with no limit. Looking at a concrete example like that may help your intuition and give you a better idea of what’s going on in this situation, but it doesn’t prove anything. Use the fact that $\langle x_n:n\in\Bbb N\rangle$ has no limit to show that each point of $X$ has an open nbhd that contains at most one point of the set $\{x_n:n\in\Bbb N\}$. May 18, 2021 at 6:15

Fix a metric on $$X$$. Let $$\mathcal U=\{U_i\}_{i\in I}$$ be an open cover of $$X'$$ that does not allow a finite subcover. Define a sequence recursively as follows:

Assume for $$n\in\Bbb N$$, we already have $$x_i\in X'$$ and $$r_i>0$$, $$1\le i such that the $$r_i$$-ball around $$x_i$$ contains no other $$x_j$$ and that the open cover $$\mathcal U$$ of $$X_n:=X'\setminus\bigcup_{1\le i does not allow a finite subcover. Pick $$x_n\in X_n$$ (possible because $$X_n=\emptyset$$ would imply a finite subcover). Pick $$U_n\in \mathcal U_n$$ with $$x_n\in U_n$$. Pick $$r_n>0$$ with $$B_{2r_n}(x_n)\subseteq U_n$$ and $$r_n for $$1\le i. Then for $$1\le i, $$x_n$$ is not in any $$B_{r_i}(x_i)$$ nor is is any $$x_i\in B_{r_n})(x_n)$$, as desired. Also, any finite subcover of $$X_{n+1}$$ could be extended to a finite subcover of $$X_n$$ by adding $$U_n$$. Hence the recursion works, and we obtain an infinte sequence $$\{x_n\}_{n=1}^\infty$$, which forms an infinite discrete subspace of $$X'$$.

Note that we did not use that $$X'$$ is the set of limit points, it could be any non-compact metrizable space.

• Each singleton set $\{x_n\}$, $n\in \mathbb N$, is open in the subspace $\{x_n:n\in \mathbb N\}$, but is not open in the space $(X,d)$.
– GGI
May 18, 2021 at 10:26

In a metrisable space $$X$$, $$X$$ compact is equivalent to $$X$$ is limit point compact (every countably infinite subset of $$X$$ has a limit point in $$X$$). This is well-known and classical (so papers don't cite this as a result due to some author in some paper etc.).

So $$X$$ not compact means that $$X$$ has a countably infinite subset that has no limit point in $$X$$ and such a set is closed and discrete in any (not just metric) space.