Given a filterbasis, by picking up an point out of every element of this filterbasis, and considering the ordering on the elements of the filterbasis $u\leq v$ iff $v \subset u$ (subquestion why is it not the other way like subset v) , we get a net.
Given a net over I, we get a filterbasis by considering $ \{ x_k| k \geq i \}$,for all $i \in I $.
Can we write a isomorphism between the set of all filters and the set of all nets. Or is there a filter not derivable from a net (or vice versa).
here they claim there is a Galois connection, i cant see the pdf an am curious how that actually would be prooven, if there is no isomorphism.
More precisely, if $\mathcal{F}$ is a filterbase on $X$, then on the directed set
$$I(\mathcal{F}) = \{(x,F): x \in F \in \mathcal{F}\}; (x_1,F_1) \le (x_2, F_2) \iff F_2 \subseteq F_1$$
we have a net $\Phi(\mathcal{F}); I(\mathcal{F}) \to X; (x,F) \to x$. (so we consider all choices for points from $F$ simultaneously, so no AC is needed in the definition, it's all canonical).
And indeed we have an inverse by using tails:
If $f: I \to X$ is a net defined on some directed set $I$, we define
$$\mathcal{F} = \{ \{f(i): i \in I, i \le i_0\} \mid i_0 \in I\} $$
the tail filter $\Psi(f)$ defined by $f$. And indeed these maps are each other's inverse (see this link (thx to Brian) for complete proofs).
This correspondence is "Galois" in the order reversal properties : a subnet of a net corresponds (under this map) to a larger filterbase, and vice versa, a larger filter(base) makes for a subnet.