Net based on finer filter is a subnet? Let $G$ be a filter finer than the filter $F$. Is it necessary that the net based on the filter $G$ is a subnet of the net based on $F$?
 A: Well, there are at least two nets that I could think of that derive from some filter on $X$: $\mathcal{F}$: one that used choices: consider $\mathcal{F}$, directed by $\supseteq$, as the index set, and for each $F \in \mathcal{F}$, pick some $\lambda(F) \in F$, and then $\lambda : \mathcal{F} \rightarrow X$ is a net based on $\mathcal{F}$. 
If we then have a finer filter $\mathcal{G}$, and we choose the two nets (one based on $\mathcal{F}$ and the other on $\mathcal{G}$) in this way independently, then there is no reason that the net based on $\mathcal{G}$ would be a subnet of the one based on $\mathcal{F}$, because this would imply that for a lot of elements they would have made the same choice (if we follows the usual definition of subnet, using cofinal maps between index sets).
A more canonical way to derive a net from a filter, is to consider the large index set 
$$I(\mathcal{F})=\left\{(F,x): F \in \mathcal{F}, x \in F\right\} $$ ordered by $(F,x) \le (G,x')$ iff $G \subseteq F$. The net derived from $\mathcal{F}$ is then $f_{\mathcal{F}}((F,x)) = x$ for $(F,x) \in I(\mathcal{F})$.
If we have a finer filter $\mathcal{G}$ and a corresponding $f_{\mathcal{G}}$, then being a subnet would mean there is a map $h$ from $I(\mathcal{G})$ to $I(\mathcal{F})$ that preserves the two orders, and such that its image is cofinal, and for all $G \in \mathcal{G}$ we have that $h((G,x))$ is of the form $(F,x)$ (same $x$), for some $F \in \mathcal{F}$. I don't really see how to do that in general (think of $\mathcal{F}$ as the cofinite filter on a set, and $\mathcal{G}$ as any free ultrafilter, e.g.).
It seems that this does not quite hold, I think, but I have no concrete counterexample. Or you use another notion of subnet (like AA-subnets).
