Let $X$ be a metric space and $A \subseteq X$. If there is a net $(x_d)_{d\in D}$ in $A$ that converges to $a \in X$, then there is a sequence $(y_n)_{n\in \mathbb N}$ in $A$ that converges to $a$. So for metric spaces, we can replace net with sequence in below results. Let $X$ be a topological ...

There is a reasonably direct proof of this fact. The trick is to take a subnet that is a sequence. Consider a nonempty net $(x_d)_{d \in D}$. If there is a maximal $d$, then the singleton net $x_d$ converges to $d$. Otherwise, for every $d$, there is some greater $d’$. Using dependent choice, cho...

Yes, it is well-known that Let $X$ be a sequential space. Then $X$ is countably compact iff $X$ is sequentially compact. To be clear about definitions: $X$ is countably compact iff every countable open cover of $X$ has a finite subbcover. I'll use the convenient equivalence that $X$ is countabl...