From Engelking's book on general topology (slightly rephrased):

Definition: We say that the net $S': \Sigma' \to X$ is finer than the net $S: \Sigma \to X$ if
1. there exists a function $f: \Sigma' \to \Sigma$ such that for any $\sigma'_0 \in \Sigma'$ there exists a $\sigma_0 \in \Sigma$ such that for any $\sigma' \in \Sigma'$ $\sigma'_0 \leq \sigma'$ implies $\sigma_0 \leq f(\sigma')$,
2. $S \circ f = S'$. We call such an $f$ a refinement.

Now the most obvious definition of a net morphism is a preorder morphism of the underlying directed sets satisfying (2). My question: is there any need to consider refinements which are not morphisms?

To be more precise:

Conjecture: For any refinement $f$ between $S': \Sigma' \to X$ and $S: \Sigma \to X$ there exists a morphism $\hat f: S' \to S$ such that $\hat f$ is a refinement, and $\hat f(\Sigma') \subset f(\Sigma')$.

In plain words, this means that any refinement can be refined further using a morphism from the same net.

I can't quite get a knack of how to prove this conjecture or construct a counterexample. Also it may be that I'm overcomplicating matters and refinements can be reduced to refinement morphisms in a simpler manner. Any tips?

  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Subnet_(mathematics): there they use "your" definition (cofinal monotone set; cofinal is necessary though); one can show (I think Schechter's book does) that these notions yield the same basic theorems wrt subnets (compact iff every net has a convergent subnet etc.). So the extra generality does not give extra results, but can be more convenient (easier to construct). Maybe the proof of this equivalence has a result of this kind? $\endgroup$ Oct 13, 2011 at 13:03
  • $\begingroup$ You may want to look at Pete Clark's notes on convergence. $\endgroup$
    – t.b.
    Oct 13, 2011 at 18:18
  • $\begingroup$ Talk page of that wikipedia article discusses several possible definitions of subnet (=finer net). Schechter's book is also mentioned there. $\endgroup$ Nov 23, 2011 at 17:49


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