Subnets and finer filters Suppose $G$ is a finer filter than $F$ in a topological space $X$. 
Is the net base in $G$ a subnet of the net base in $F$?
I am using the definitions of General Topology of Willard:

Definition 12.15. If $(x_\lambda)$ is a net in $X$, the filter generated by the filter base $\mathscr C$ consisting of the sets $B_{\lambda_0}=\{x_\lambda; \lambda\ge\lambda_0\}$, $\lambda_0\in\Lambda$, is called the filter generated by $(x_\lambda)$.
Definition 12.16. If $\mathscr F$ is a filter on $X$, let $\Lambda_{\mathscr F}=\{(x,F); x\in F\in\mathscr F\}$.
  Then $\Lambda_{\mathscr F}$ is directed by the relation $(x_1,F_1)\le(x_2,F_2)$ iff $F_2\subset F_1$, so the map
  $P\colon \Lambda_{\mathscr F} \to X$ defined by $P(x,F)=x$ is a net in $X$. It is called the net based on $\mathscr F$.

The definition of subnet used in Willard is the second one from this post: Different definitions of subnet
Thank you
 A: Not necessarily. I’ll start by adding the relevant definitions. 
First, your terminology isn’t quite right: you mean the net based on $\mathscr G$ or $\mathscr F$, not the net base in. The net $\nu_{\mathscr{G}}$ based on $\mathscr G$ is defined as follows. Let $\Lambda_{\mathscr{G}}=\{\langle x,G\rangle:x\in G\in\mathscr{G}\}$, and partially order $\Lambda_{\mathscr{G}}$ by setting $\langle x_0,G_0\rangle\preceq_{\mathscr{G}}\langle x_1,G_1\rangle$ if and only if $G_0\supseteq G_1$. Then $\nu_{\mathscr{G}}:\Lambda_{\mathscr{G}}\to X:\langle x,G\rangle\mapsto x$. The net $\nu_{\mathscr{F}}$ based on $\mathscr{F}$ is defined similarly.
In Willard’s terminology $\nu_{\mathscr{G}}$ is a subnet of $\nu_{\mathscr{F}}$ if and only if there is an increasing cofinal map $\varphi:\Lambda_{\mathscr{G}}\to\Lambda_{\mathscr{F}}$ such that $\nu_{\mathscr{G}}=\nu_{\mathscr{F}}\circ\varphi$.
Let $\mathscr{F}=\{F\subseteq\Bbb N:0\in F\text{ and }\Bbb N\setminus F\text{ is finite}\}$, and let $\mathscr{G}=\{G\subseteq\Bbb N:0\in G\}$; clearly $\mathscr{F}$ and $\mathscr{G}$ are filters on $\Bbb N$, and $\mathscr{G}$ is finer than $\mathscr{F}$. Suppose that $\varphi:\Lambda_{\mathscr{G}}\to\Lambda_{\mathscr{F}}$ is increasing. Let $\langle n,F\rangle=\varphi(\langle 0,\{0\}\rangle)$, and note that $\langle 0,\{0\}\rangle$ is the maximum element of $\Lambda_{\mathscr{G}}$. Fix $m\in F\setminus\{0,n\}$, and let $F'=F\setminus\{m\}$; clearly $\langle n,F'\rangle\in\Lambda_{\mathscr{F}}$, and $\langle n,F\rangle\precneqq_{\mathscr{F}}\langle n,F'\rangle$. Suppose that $\langle k,G\rangle\in\Lambda_{\mathscr{G}}$ satisfies $\langle n,F'\rangle\preceq_{\mathscr{F}}\varphi(\langle k,G\rangle)$; then $\langle k,G\rangle\preceq_{\mathscr{G}}\langle 0,\{0\}\rangle$, but 
$$\varphi(\langle 0,\{0\}\rangle)=\langle n,F\rangle\precneqq_{\mathscr{F}}\langle n,F'\rangle\preceq_{\mathscr{F}}\varphi(\langle k,G\rangle)\;,$$
and therefore $\varphi(\langle k,G\rangle)\npreceq_{\mathscr{F}}\varphi(\langle 0,\{0\}\rangle)$, contradicting the hypothesis that $\varphi$ is increasing. Thus, for all $\langle k,G\rangle\in\Lambda_{\mathscr{G}}$ we must have $\langle n,F'\rangle\npreceq_{\mathscr{F}}\varphi(\langle k,G\rangle)$, and $\varphi$ therefore does not map $\Lambda_{\mathscr{F}}$ cofinally into $\Lambda_{\mathscr{G}}$. It follows immediately that $\nu_{\mathscr{G}}$ is not a subnet of $\nu_{\mathscr{F}}$.
