5
$\begingroup$

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 $n$?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ It is worth mentioning that there are other possibilities to assign a net to a filter. It is possible to do it in a such way that finer filter gives a subnet. This is discussed in Pete L. Clarks' notes on general topology. I will also mention that there exist several definitions of subnet. The AA-subnet is defined in such a way, that finer filter corresponds to a subnet; but for this definition of subnet this fact is rather trivial. $\endgroup$ Commented Apr 14, 2014 at 15:34
  • $\begingroup$ @MartinSleziak Thank you for the references (+1). Your first link seems defunct. $\endgroup$ Commented Feb 20, 2019 at 7:46
  • 1
    $\begingroup$ @DuchampGérardH.E. It seems that Pete L. Clark moved the site here: alpha.math.uga.edu/~pete/expositions2012.html (Wayback Machine). The current link to the notes on general topology is math.uga.edu/~pete/pointset2018.pdf (Wayback Machine). $\endgroup$ Commented Feb 20, 2019 at 8:01
  • 1
    $\begingroup$ I'll include also Wayback Machine link for the second link. (I should have done this when I first posted this, I do apologize.) $\endgroup$ Commented Feb 20, 2019 at 12:06
  • $\begingroup$ @MartinSleziak Thanks (+1+1) $\endgroup$ Commented Feb 22, 2019 at 7:22

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.