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

  • 1
    $\begingroup$ It's good that you try to tell us which definitions you use, but for those of us who haven't read Willard's book, saying that you use its definitions unfortunately doesn't help. What is "the net base in $G$"? $\endgroup$ – Daniel Fischer May 20 '14 at 13:49
  • $\begingroup$ Perhaps it is worth mentioning that this is not the only assign filter to a given net in a "reasonable" way. (Reasonable meaning that it preserves some properties, such as convergence, cluster points, etc.) There is also a correspondence which will preserve finer filter/subnets in both direction (unlike this one), although the definition of the net corresponding to a given filter becomes more complicated. You can read more about this in Pete L. Clark's notes on convergence. $\endgroup$ – Martin Sleziak Jan 12 '16 at 7:59

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 $\nu_{\mathscr{F}}$.


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