I am trying to solve the following problem:

Let $P : \Lambda \to X$ be a net in a topological space $X$, and let $Q = P \circ \phi : M \to X$ be a subnet of $P$ (i.e. $\Lambda$ and $M$ are directed sets, $\phi : M \to \Lambda$, and $(\forall \lambda_0 \in \Lambda)(\exists \mu_0 \in M)(\forall \mu \in M)(\mu \geq \mu_0 \Rightarrow \phi(\mu) \geq \lambda_0)$).

I want to construct a subnet $R = Q \circ \psi : \Sigma \to X$ of $Q$ such that the function $\phi \circ \psi : \Sigma \to \Lambda$ is non-decreasing, i.e. $\sigma \leq \sigma' \Rightarrow (\phi \circ \psi)(\sigma) \leq (\phi \circ \psi)(\sigma')$.

The hint that I have is to choose $\Sigma$ to be a subset of $M \times \Lambda$ or $\Lambda \times M$. So far I have tried the following idea:

$\Sigma = \{ (\phi(\mu'), \mu) \in \Lambda \times M \space | \space \mu' \in M \space \& \space (\forall \mu_0 \geq \mu)(\phi(\mu_0) \geq \phi(\mu')) \}$, with the direction given by $(\phi(\mu'), \mu) \leq (\phi(\nu'), \nu) \Leftrightarrow \mu \leq \nu \space \& \space \phi(\mu') \leq \phi(\nu')$.

Then I have shown that $\Sigma$ so defined is a directed set, and that if we define $\psi : \Sigma \to M$ to be $\psi(\phi(\mu'), \mu) = \mu$, then $R = Q \circ \psi : \Sigma \to X$ is a subnet of Q. However, I can't seem to prove that the function $\phi \circ \psi : \Sigma \to \Lambda$ is non-decreasing.

I would greatly appreciate any hints or answers!


HINT: For $\lambda\in\Lambda$ and $\mu\in M$ let $\Lambda_\lambda=\{\kappa\in\Lambda:\lambda\le_\Lambda\kappa\}$ and $M_\mu=\{\nu\in M:\mu\le_M\nu\}$. Let

$$\Sigma=\{\langle\lambda,\mu\rangle\in\Lambda\times M:\varphi[M_\mu]\subseteq\Lambda_\lambda\}\;.$$

For $\langle\lambda,\mu\rangle,\langle\kappa,\nu\rangle\in\Sigma$ let $\langle\lambda,\mu\rangle\preceq\langle\kappa,\nu\rangle$ if and only if $\varphi(\mu)\le_\Lambda\kappa$ and $\mu \leq \nu$, or $\langle\lambda,\mu\rangle=\langle\kappa,\nu\rangle$. Finally, define

$$\psi:\Sigma\to M:\langle\lambda,\mu\rangle\mapsto\mu\;.$$

  • $\begingroup$ The direction as you have defined it does not seem to be reflexive. So I think I will have to modify it a bit. $\endgroup$ – User7819 Nov 13 '15 at 17:57
  • 1
    $\begingroup$ @Jason: You’re right. Taking its reflexive closure appears to do the trick, though, and I’ve modified the definition accordingly. $\endgroup$ – Brian M. Scott Nov 13 '15 at 20:11
  • $\begingroup$ So with this modification, we now have a directed set, although we do not seem to have a subnet. $\endgroup$ – User7819 Nov 16 '15 at 2:40
  • $\begingroup$ The problem seems to be that the definition of the direction does not explicitly involve the second component of the second pair. $\endgroup$ – User7819 Nov 16 '15 at 2:43
  • $\begingroup$ But I added an edit that seems to fix the problem. $\endgroup$ – User7819 Nov 16 '15 at 2:53

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