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

Question

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.

Answer 1

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 $X$.

Answer 2

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 $r_i>0$, $1\le i<n$ such that the $r_i$-ball around $x_i$ contains no other $x_j$ and that the open cover $\mathcal U$ of $X_n:=X'\setminus\bigcup_{1\le i<n}\overline{B_{r_i}(x_i)}$ 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 $r_n<r_i$ for $1\le i<n$. Then for $1\le i<n$, $x_n$ 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.

Answer 3

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.