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Let $(D, \geq)$ be a directed set and let $(n_d)_{d\in D}$ be a real-valued net. Assume $a\in \mathbb R$ is a cluster point of $(n_d)$, i.e., for every neighborhood $U$ of $a$ and every $d\in D$ there exists $d'\geq d$ such that $n_{d'}\in U$.

It is a standard fact that there exists a subnet of $(n_d)$ that converges to $a$.

But is it possible to find a cofinal subnet (see for example wikipedia) with the same property? To rephrase: is there a cofinal subset $D'\subset D$ such that $\lim_{d\in D'}n_d = a$?

If $(n_d)$ takes value in an arbitrary topological space $X$, then this is not true. See for example this answer and this one. My question is specific to the case $X=\mathbb R$, and for this space I could not find any counter-example nor proof.

Navukido20
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1 Answers1

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Not necessarily.

Let $D=\omega_1\times\omega$ ordered lexicographically: $\langle\alpha,m\rangle\preceq\langle\beta,n\rangle$ iff $\alpha<\beta$, or $\alpha=\beta$ and $m\le n$; this is clearly a directed set. Define a net

$$\nu:D\to\Bbb R:\langle\alpha,n\rangle\mapsto 2^{-n}\,;$$

clearly $0$ is a cluster point of $\nu$.

Let $C$ be a cofinal subset of $D$. For each $n\in\omega$ let $C_n=C\cap(\omega_1\times\{n\})$; $|C|=\omega_1$, so there is some $m\in\omega$ such that $|C_m|=\omega_1$. Then $C_m$ is cofinal in $C$, but $\nu[C_m]=\{2^{-m}\}$ is disjoint from the nbhd $(-2^{-m},2^{-m})$ of $0$. Thus, no cofinal subnet of $\nu$ converges to $0$.

Brian M. Scott
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  • Excellent, thank you very much! – Navukido20 Feb 16 '21 at 20:24
  • @Navukido20: You’re very welcome. – Brian M. Scott Feb 16 '21 at 20:28
  • After thinking about it a bit more, I wonder if the following additional assumption on the directed set $D$, which fails in your counter-example, would suffice to make the conclusion hold: assume there exists a real-valued net $(r_d){d\in D}$ such that $\lim{d\in D}r_d = +\infty$. [Please let me know if it is unappropriate to ask such a question in the comments...] – Navukido20 Feb 16 '21 at 21:50
  • @Navukido20: Let me think about it for a day or so; if I’m not getting anywhere, I’ll let you know, and you can post it as a new question. – Brian M. Scott Feb 17 '21 at 02:11
  • Much appreciated, many thanks! – Navukido20 Feb 17 '21 at 12:57
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    @Navukido20: The condition that you suggest is at least a necessary condition: it’s equivalent to requiring that there exist a $D$-net converging to some real number, which in turn is equivalent to requiring that each (extended) real number be the limit of some $D$-net. It implies that $D$ has an unbounded increasing sequence, unlike the directed set of my example, but I’ve not had as much time to think about it as I’d hoped, and I’ve not taken it further than that yet. – Brian M. Scott Feb 18 '21 at 18:03
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    Thanks again! That is a very good point, the additional assumption is equivalent to $D$ having an unbounded increasing sequence, which is a nicer formulation and indeed a necessary condition! So now it remains to determine if it is sufficient... I'd be very happy if you have any idea or hint about this, and otherwise I'll post it as a separate question in a couple of days. – Navukido20 Feb 19 '21 at 07:37