Prove a map $\varphi$ between all nets and all filters is surjective.

Question

We define such a map:

$\varphi(N) = \{A \subset X \ | \ \exists \alpha \in \Omega, \forall \beta \ge \alpha: f(\beta) \in A \}$

Here $N = (\Omega, f, \ge)$ is a net with index set $\Omega$.

This way $\varphi$ is a map between the set of all nets on $X$ and the set of all filters on $X$.

I'm not able to prove that this map is surjective. In other words, that for every filter $\mathcal A$ on $X$ there is such net $N'$ that $\varphi(N') = \mathcal A$.

Does anyone have an idea how to prove that?

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By constructing such a net $N' = (\mathcal A, f, \subseteq)$ for a given $\mathcal A$, where $f$ is a map that for each $A \in \mathcal A$ it returns some element $a \in A$.

I can prove that $\mathcal A \subset \varphi(N') = \{A \subset X \ | \ \exists \alpha \in \mathcal A, \forall \beta \in \mathcal A, \beta \subseteq \alpha: f(\beta) \in A \}$ but not the reverse direction. Does the reverse direction hold in such case, if so then why?

Answer 1

The clue is, make no choices but allow all choices.

Given $\mathcal{A}$ define

• $\Omega = \{(A,a)\mid A \in \mathcal{F}, a \in A\}$.
• $(A,a) \ge (B,b)$ iff $A \subseteq B$ (this together defines the directed set as domain).
• $f((A,a))=a$ is the actual net mapping. Together (domain and mapping) I call them $N(\mathcal{A})$, say.

Then (tedious check that you should do in detail):

$\phi(N(\mathcal{A})) = \mathcal{A}$. Fact: the "tail set" for index $\alpha=(A,a)$ onwards is just $A$ again.