Let's define a net and subnet in this way:


* A net is any function of the form $n:(P,\le)\to X$ where $(P,\le)$ is a (preordered) directed set.
* A net $m:(P',\le)\to X$ is a subnet of the net $n:(P,\le)\to X$ iff there is a function:
$$\theta:(P',\le)\to (P,\le)$$
which is increasing:
$$x'\le y' \to \theta(x')\le\theta (y')$$
and cofinal:
$$(\forall p\in P)(\exists p'\in P')(p\le\theta(p'))$$
and $m=n\circ\theta$.
* The filter assigned to a net $n:(P\le)\to X$ is the filter generated by
$$\lbrace \lbrace n(x)\mid x\ge p \rbrace \mid p\in P\rbrace$$
on X. We denote this filter by $\mathcal F\_n$,


My question is:


Are there nets $m:(P',\le)\to X$ and $n:(P,\le)\to X$ with
$$\mathcal F\_n\subseteq \mathcal F\_m$$
such that $m$ is not a subnet of $n$?



Connection between subnet and superfilter
Yes. Let Z be the integers, let X consist of a single point and let m : {0} -> X and n : Z -> X be constant functions. Then n and m give the same filter but m cannot be
a subnet of n since no single integer in Z is cofinal.


