How is it proved that if every infinite subset of $E$ has at least an accumulation point in $E$, then $E$ is compact?

Question

I know how to prove that every infinite subset $A$ of a compact set in a metric space satisfies $A'\ne\emptyset$, but my book also claims the opposite implication, without proof, stating "it falls outside the scope of this book". I've been searching online but I could only find the "standard" result.

Answer 1

Assume every infinite subset of $E$ has a limit point. Then fix some $\delta > 0$, choose some $x_1 \in E$, and then given $x_1,\ldots,x_n$, choose $x_{n+1}$ such that $d(x_{n+1},x_i) \ge \delta$ for $i = 1,\ldots,n$. This terminates in finitely many steps, so $X$ can be covered in finitely many open balls of radius $\delta$. This implies that $E$ is totally bounded. Now, a sequence in $E$ must have a convergent subsequence, so in particular, a Cauchy sequence must converge. $E$ is complete and totally bounded, so it is compact.

Answer 2

We prove this by contradiction.

Suppose $\{F_n\}_{n \in \mathbb{N}}$ is an open cover of $A$ without a finite subcover (i.e. $A$ is not compact). Then define $G_n = (\bigcup_{i=1}^n F_i)^c$. For each $n$, $G_n$ must be nonempty, else that finite subcollection $F_1 \cup ... \cup F_n$ would be a finite subcover of $A$.

Now define $E = \{x_n\}_{n \in \mathbb{N}}$ such that $x_i \in G_i$. Clearly this set is infinite by what we have shown above, and by hypothesis it has a limit point $x \in A$. Then $x \in F_m$ for some m. Since $F_m$ is open, there exists an $\epsilon \gt 0$ such that $B_\epsilon(x) \subset F_m$. By how we have constructed $E$, there can only be finitely many points in $B_\epsilon(x)$, since no point $x_n$ of $E$ with $n \gt m$ is in $F_m$. But then $x$ is not a limit point of $E$, since every neighborhood of a limit point of a set must contain infinitely many elements of the set. Contradiction.

Thus there must be a finite subcover, and since $\{F_n\}_{n \in \mathbb{N}}$ was arbitrary, we can conclude $A$ is compact.

Answer 3

Let $(X,d)$ be a metric space.

(1). If $D$ is a dense subset of $X$ then $\mathbb B=\{B_d(x,q): x\in D\land q\in \mathbb Q^+\}$ is a base for $X.$ Note that if $D$ is countable then $\mathbb B$ is countable.

(2). If $X$ has a countable base then every open cover of $X$ has a countable sub-cover.

Let $(X,d)$ be a metric space such that $A'\ne \emptyset$ whenever $A$ is an infinite subset of $X.$

(i). For $q\in \mathbb Q^+$ let $S_q\subset X$ such that $\{B_d(s,q):s\in S_q\}$ is a maximal family of pair-wise disjoint open balls, where $B_d(s,q)\ne B_d(t,q)$ for distinct $s,t\in S_q$. Then $S_q$ is finite (because $S'$ is empty because $d(s,t)\geq q$ for distinct $s,s'\in S_q$).

(ii). Let $D=\cup_{q\in \mathbb Q^+}S_q$ . Then $D$ is countable.

$D$ dense in $X$. Because otherwise $B_d(x,r)\cap D=\emptyset$ for some $x\in X$ and some $r>0.$ But take $q\in \mathbb Q\cap (0,r/2).$ Then $B_d(x,q)\cap B_d(s,q)=\emptyset$ for all $s\in S_q,$ contradicting the maximality of $\{B_d(s,q):s\in S_q\}.$

By (1), let $B$ be a countable base for $X.$

(iii). Let $C$ be an open cover of $X.$ By (2), let $C'=\{c_n: n\in \mathbb N\}$ be a countable subcover.Suppose,by contradiction, that $C'$ has no finite subcover.

Then for $n\in \mathbb N$ let $x_n\in X$ \ $\cup_{j=1}^nc_j.$ For each $n$ there are only finitely many $n'$ for which $x_n=x_{n'}$ because $x_n\in c_m$ for some $m,$ and $x_{n'}\not \in c_m$ for all $n'\geq m.$

So $A=\{x_n: n\in \mathbb N\}$ is an infinite set. By hypothesis, $A$ has an accumulation point $p.$ Now $p\in c_m$ for some $m,$ so $A\cap c_m$ is infinite, so $\{n: x_n\in c_m\}$ is infinite. But this implies there exists $n>m$ with $x_n\in c_m\subset \cup_{j=1}^mc_j,$ contrary to the def'n of $x_n .$

Therefore $C'$ has a finite sub-cover.

Remark: The Q is equivalent to: A non-compact metric space has an infinite closed discrete subspace. This is useful in other q's. For example we can use it to prove that a non-compact metrizable space has an unbounded metric. Another example: The $\epsilon$-order topology on $\omega_1$ is not metrizable, because it is non-compact but has no infinite closed discrete subspace.