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In a sequence of compactness criterions of Metric Spaces, the final step is to prove that every countably compact metric space is compact, so I can't invoke sequential compactness or total boundedness etc. But I know the steps I have to take to prove this.

  1. Show that any countably compact metric space is separable
  2. A separable metric space is second-countable
  3. A second countable metric space is a Lindelöf space
  4. Any countably compact Lindelöf space is compact.

I can work through the latter 3, but I'm having trouble proving the 1st one without total boundedness. Also, can someone confirm that these steps are correct. Thanks!

PS: This is the sequence of the compactness criterions that I have, and I'm dealing with the final one:

Compactness $\implies$ Sequential Compactness $\implies$ Complete and Totally Bounded $\implies$ Limit Point Compactness $\implies$ Finite Intersection Property $\implies$ Countable Compactness $\implies$ Compactness

PS: I know there have been some similar questions, but those ask to invoke equivalent definitions that I can't, as I've mentioned in the question.

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  • $\begingroup$ In a countably compact metric space every infinite subset has a limit point. Thus every set whose elements have a distance $\geqslant 1/n$ from each other must be finite. $\endgroup$ – Daniel Fischer Mar 24 '18 at 18:30
  • $\begingroup$ Sorry, can't use limit point compactness either. That's part of the sequence of equivalent definitions $\endgroup$ – Naweed G. Seldon Mar 24 '18 at 18:31
  • $\begingroup$ Finite intersection property in its general form directly implies compactness. $\endgroup$ – Henno Brandsma Mar 24 '18 at 22:17
  • $\begingroup$ @HennoBrandsma Yes, I know. But that's not the question I have, unfortunately $\endgroup$ – Naweed G. Seldon Mar 24 '18 at 22:18
  • $\begingroup$ @DanielFischer if you take a maximal such set it's closed and discrete and this is enough to contradict countable compactness. $\endgroup$ – Henno Brandsma Mar 25 '18 at 15:36
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Lemma: Let $(X,d)$ be a metric space. For each $n \in \{1,2,3,\ldots\}$ there is (by Zorn's lemma) a maximal (by inclusion) set $D_n$ such that for all $x,y \in D_n$ with $x \neq y$, we have $d(x,y) \ge \frac{1}{n}$. If all $D_n$ are at most countable, then $X$ is separable.

Proof: We'll show that $D = \cup_n D_n$ is dense (and is countable when all $D_n$ are), and so pick any $x \in X$ and any $r>0$ and we'll show that $B(x,r) \cap D \neq \emptyset$. Suppose not, then find $m$ with $\frac{1}{m} < r$. Then from $B(x,r) \cap D = \emptyset$ we know that $d(x, y) \ge \frac{1}{m}$ for all $y \in D_m$ (or else $y \in D \cap B(x,r)$), and then $D_m \cup \{x\}$ would contradict the maximality of $D_m$. So the intersection with $D_m$ and hence $D$ is non-empty, and so $D$ is dense.

(The above is the heart of the proof that a ccc metric space is separable, as $X$ ccc implies that all $D_m$ are countable, as the $B(x,\frac{1}{m})$, $x \in D_m$ are a pairwise disjoint open family of non-empty sets.)

Now suppose that $(X,d)$ is countably compact. Suppose for a contradiction that $X$ is not separable. Then for some $m \ge 1$ we have that $D_m$ (as in the lemma) is uncountable. Fix such an $m$.

Now $D_m$ is discrete (clear as $B(x,\frac{1}{m}) \cap D_m = \{x\}$ for each $x \in D_m$) and closed: suppose that $y \in X\setminus D_m$ is in $\overline{D_m}$, then $B(y, \frac{1}{2m})$ contains infinitely many points of $D_m$ and for any $2$ of them, say $x_1, x_2 \in D_m$, we'd have $d(x_1, x_2) \le d(x_1, y) + d(y,x_2) < \frac{1}{m}$ contradiction (as points in $D_m$ are at least $\frac{1}{m}$ apart). So $D_m$ is closed and discrete (as an aside: being uncountable this would already contradict Lindelöfness of $X$; this shows that a Lindelöf metric space is separable, e.g.), but we want to contradict countable compactness, which is easy too:

Choose $A \subseteq D_m$ countably infinite (so that $A$ is closed in $X$, as all subsets of $D_m$ are), and define a countable open cover $\mathcal{U} = \{B(p, \frac{1}{m}): p \in A\} \cup \{X\setminus A\}$ of $X$ that has no finite subcover: we need every $B(p,\frac{1}{m})$ to cover $p$ for all $p \in A$.

This contradiction then shows $X$ is separable (and thus has a countable base ,is Lindelöf etc. finishing the compactness).

The crucial fact is that all of the following are equivalent for a metric space:

  1. $X$ has a countable base.
  2. $X$ is separable.
  3. All discrete subspaces of $X$ are at most countable.
  4. All closed and discrete subspaces of $X$ are at most countable.
  5. $X$ is ccc.
  6. $X$ is Lindelöf.

In the above I essentially did "not (2) implies not (4)" implicitly. The fun is that countable compactness implies (4) easily (such closed discrete subspaces are even finite) and thus we get Lindelöfness "for free". Also implicit in the above proof is that every countably compact space is limit point compact.

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