Given a directed set, how do I construct a net with positive values that converges to $0$.

Question

Suppose I have a directed set in which there is no maximal or equivalently maximum element. Is there a way to construct a net based on that directed set, of strictly positive real numbers that converges to $0$? This would be useful in arguments where for sequences one uses $1/n$.

Answer 1

Let $\langle D,\le\rangle$ be a directed set. Suppose that $\nu:D\to\Bbb R$ is a net based on $D$ such that $\nu_d>0$ for each $d\in D$ and $\nu\to 0$. Choose $d_0\in D$ so that $\nu_d<1$ whenever $d\ge d_0$. Given $d_n$, choose $d_{n+1}\in D$ so that $d_{n+1}>d_n$ and $\nu_d<2^{-(n+1)}$ whenever $d\ge d_{n+1}$. This construction is possible because $\nu\to 0$.

Suppose that there were some $e\in D$ such that $d_n\le e$ for every $n\in\Bbb N$; then clearly we’d have $0<\nu_e<2^{-n}$ for every $n\in\Bbb N$, which is absurd. Thus, no such $e$ can exist: the set $\{d_n:n\in\Bbb N\}$ is unbounded in $D$.

Conversely, suppose that the directed set $D$ has an unbounded sequence $\langle d_n:n\in\Bbb N\rangle$. Without loss of generality we may assume that $d_0\le d_1\le\dots\;$. Fix $d\in D$; there is some $m\in\Bbb N$ such that $d_m\not\le d$. (Otherwise the sequence $\langle d_n:n\in\Bbb N\rangle$ would be bounded by $d$.) Let $m(d)=\min\{n\in\Bbb N:d_n\not\le d\}$, and let $\nu_d=2^{-m(d)}$; this defines the net $\nu:D\to\Bbb R$, and clearly $\nu_d>0$ for each $d\in D$. If $\epsilon>0$, choose $n\in\Bbb N$ so that $2^{-n}<\epsilon$. Now suppose that $d\ge d_n$; then $m(d)>n$, so $\nu_d<2^{-n}<\epsilon$. Thus, $\nu\to 0$.

This shows that a necessary and sufficient condition for the existence of the kind of net that you want is that the directed set $D$ contain an unbounded sequence. The first uncountable ordinal, $\omega_1$, is not such a directed set, because every countable subset of $\omega_1$ is bounded.