A direct proof (using net convergence) that sequential compact metric space is compact

Question

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 space and $A \subseteq X$. Then $\overline A$ is the set of limits of convergent nets with values in $A$.

• Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be topological spaces and $f:X \to Y$ be continuous, i.e., $f^{-1} (O) \in \tau_X$ for all $O \in \tau_Y$. Then $f$ is continuous if and only if, $f(x_d) \to f(x)$ where $(x_d)_{d \in D}$ is a net such that $x_d \to x$.

Usually, the proof of sequential compactness implying compactness is indirect (by contradiction), uses the cover definition, and involves auxiliary lemmas. Is there a direct proof that uses below net definition?

Let $X$ be a topological space. Then $X$ is compact if and only if every net in $X$ has a convergent subnet.

Answer 1

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, choose $d_1 < d_2 < \ldots$. Then $(x_{d_n})_{n \in \mathbb{N}}$ has a convergent subsequence which is also a convergent subnet of the original net.

If you wish for a constructive proof that every sequentially compact space is compact, I’m afraid you’re going to be disappointed (though I haven’t ruled out the possibility that there’s a constructive proof that sequentially compact implies net compact).

We can show constructively that every compact space $X$ either has a point or is empty. For consider the open cover $\{X \mid \exists y (y \in X)\}$. This open cover has a finitely enumerator sub cover $U_1, \ldots, U_n$. If $n = 0$, then $X$ Is empty, while if $n > 0$, then $X$ has a point.

Now for any proposition $p$, consider the associated topological space $\{0 \mid p\}$. This space is sequentially compact (in fact, net compact) and has a point if and only if $p$.

If we could show this space is also compact constructively, then we would have proved the space is either empty or no empty, and thus $p \lor \neg p$. But this cannot be done constructively. So no constructive proof can exist.