Countably Compact Metric Space is Compact
Theorem
Let $M = \left({A, d}\right)$ be a metric space.
Let $M$ be countably compact.
Then $M$ is compact.
Proof 1
This follows directly from:
- Metric Space is First-Countable
- Countably Compact First-Countable Space is Sequentially Compact
- Sequentially Compact Metric Space is Second-Countable
- Second-Countable Space is Compact iff Countably Compact
$\blacksquare$
Proof 2
We have that a Metric Space is Countably Compact iff Sequentially Compact.
Then we have that a Sequentially Compact Metric Space is Separable.
For each $n$, a metric space which is countably compact can be covered by finitely many open $\left({1/n}\right)$-balls: $B_{1/n} \left({x_i}\right)$.
why?
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So $\left\{{x_i}\right\}$ is a dense subset of $A$ which is countable.
So if a metric space is countably compact it is by definition second-countable.
by definition? See unresolved issues in Talk page.
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The result follows from Second-Countable Space is Compact iff Countably Compact.
$\blacksquare$
Axiom of Countable Choice
This theorem depends on the Axiom of Countable Choice.
Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.
As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.
