Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
4  
It depends on the directed set. If your directed set is $\omega_1$ with the usual ordering, for instance, it can’t be done: any net based on that directed set that converges to $0$ in the reals is necessarily eventually constant at $0$. –  Brian M. Scott Aug 30 '12 at 5:59
    
@BrianM.Scott can you tell me what $\omega_1$ is? Also, given that this doesn't work, the real project I was doing is trying to prove that the two usual definitions of limsup for sequences generalize to nets. That is the sup of all subnet limits (subnets are compositions with cofinal increasing maps coming out of a directed set) versus the limit of the sup of the tail ends. Existence of both quantities as an extended real element is easy, and $LHS \leq RHS$ is easy. I feel that to obtain the other inequality I would need the fact above, so can I do without it? –  Jeff Aug 30 '12 at 6:05
    
I actually just found out what $\omega_1$ is, and am now more interested in why your statement holds. So every ordinal is either the initial 0, a limit, or a successor. Clearly then $\omega_1$ is a limit, so it's the sup of all the countable ordinals. I suspect some counting argument must work now, but I'm just not seeing it. –  Jeff Aug 30 '12 at 6:14
1  
Suppose that the net $\langle x_\xi:\xi\in\omega_1\rangle$ converges to $0$, where each $x_\xi>0$. Then for each $n\in\Bbb N$, there is a $\xi_n\in\omega_1$ such that $x_\xi<2^{-n}$ whenever $\xi\ge\xi_n$. Let $\eta=\sup_n\xi_n$; then $\eta$ is still a countable ordinal, so $\eta\in\omega_1$. But if $\xi\ge\eta$, then $\xi\ge\xi_n$ for every $n$, so $x_\xi<2^{-n}$ for every $n$, and therefore $x_\xi\le 0$, a contradiction. –  Brian M. Scott Aug 30 '12 at 6:17

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.