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The limsup on sequences of extended real numbers is usually taken to be either of these two things, which are equivalent:

  1. the sup of all subsequential limits.
  2. The limit of the sup of the tail ends of the sequence.

For the situation with nets, the same arguments guarantee the existence of the above quantities 1. and 2, as long as you understand that a subnet of a net must be precomposed with an increasing function that is also cofinal. Also, one has that 1. $\leq$ 2., and I just can't see the reverse inequality. Don't forget that one cannot imitate the argument for sequences because the following fact fails:

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

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Jeff
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1 Answers1

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As far as I can say, the more usual definition of limit superior of a net is the one using limit of suprema of tails: $$\limsup x_d = \lim_{d\in D} \sup_{e\ge d} x_e = \inf_{d\in D} \sup_{e\ge d} x_e.$$

But you would get an equivalent definition, if you defined $\limsup x_d$ as the largest cluster point of the net. This definition corresponds (in a sense) to the definition with subsequential limits, since a real number is a cluster point of a net if and only if there is a subnet converging to this number.

I think that it is relatively easy to see that $\limsup x_d$ is a cluster point of the net $(x_d)_{d\in D}$.

To see that for every cluster point $x$ we have $x\le\limsup x_d$ it suffices to notice that, for any given $\varepsilon>0$ and $d\in D$, the interval $(x-\varepsilon,x+\varepsilon)$ must contain some element $x_e$ for $e\ge d$. Hence we get $$ \begin{align*} x-\varepsilon &\le \sup_{e\ge d} x_e\\ x-\varepsilon &\le \lim_{d\in D} \sup_{e\ge d} x_e. \end{align*}$$ and, since $\varepsilon>0$ is arbitrary, we get $$x\le \lim \sup_{e\ge d} x_e.$$

Thus the limit superior is indeed the maximal cluster point.

So the only thing missing is to show that cluster points are precisely the limits of subnets - this is a standard result, which you can find in many textbooks.

Some references for limit superior of a net are given in the Wikipedia article and in my answer here.

Perhaps some details given in my notes here can be useful, too. (The notes are still unfinished.) I should mention, that I pay more attention there to the notion of limit superior along a filter (you can find this in literature defined for filter base, which leads basically to the same thing). The limit superior of a net can be considered a special case, if we use the section filter; which is the filter generated by the base $\mathcal B(D)=\{D_a; a\in D\}$, where $D_a$ is the upper section $D_a=\{d\in D; d\ge a\}$.

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Martin Sleziak
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    I've taken a look at your notes, they are great. I think this post at MO might be useful for the history of filters. – Michael Greinecker Aug 30 '12 at 10:51
  • @Martin Sleziak but a finite directed set, or in any case one with a maximum element, will still have a limsup, even though the set of cluster points of the net could be empty. (I assume here you mean by cluster points of a net the set of limit points of a subset of a topological space, and that subset is the image of the net.) I have not carefully digested everything in your answer yet, but perhaps what you mean is something like the sup of the closure in the extended reals of the image of the net? I think it must be important to be careful about this since apparently the distinction... – Jeff Aug 31 '12 at 08:11
  • ... of whatever notion you intend here and subnet limits matters. I am still surprised that the subnet limit definition is not equal to the sup of the tail ends. Does anybody have an example of when this can occur? – Jeff Aug 31 '12 at 08:12
  • @Jeff I don't want to overfill the main site with too many comments, so perhaps the discussion about this could continue in chat. Let me just say here, that if $D$ has a maximal element $m$, then the limit, limsup and the unique cluster point of the net $(x_d)$ is $x_m$. – Martin Sleziak Aug 31 '12 at 09:02
  • For future readers, I want to fill in the "relatively easy" gap that the limsup is itself a cluster point of $(x_d)$. For each $d$ let $y_d$ be the sup of the $x_e$'s with $e\geq d$, so that $y_d\rightarrow y$ where $y$ is the limsup. We define a subsequence $(x_{d_n})$ converging to $y$ as follows: inductively define $d_1\leq d_2\leq d_3\cdots$ such that $y_{d_n}-1/n \leq x_{d_{n+1}} \leq y_{d_n}$. Then it is clear that $x_{d_n}\rightarrow y$ as $n\rightarrow \infty$. It's strange that this is achievable with a sequence --- have I made a mistake? – Ehsaan May 01 '20 at 15:48
  • @Ehsaan I don't see how would you argue in the proof you suggested that $(x_{d_n})$ is a subset of $(x_d)$. Anyway, to show that $y$ is a cluster point of the given net it is enough to show that if $U$ is a neighborhood of $y$ then the set ${d\in X; x_d\in U}$ is cofinal in $D$. – Martin Sleziak May 01 '20 at 15:59
  • I see --- the $d_n$'s may not be cofinal. Makes sense. – Ehsaan May 01 '20 at 16:01
  • I'm still not convinced. I can see that every cluster point is a limit of a subnet, but that's where you take the definition of "subnet" to simply mean a cofinal function between the indexing sets. If you want an increasing cofinal function as in the wikipedia article you linked, I'm not sure how to do it. (This is the difference between "Kelley" and "Willard" subnets.) – Ehsaan May 02 '20 at 19:56
  • Sorry to keep this comment chain going, but I found my problem: a subnet $(x_{f(j)}){j\in J}$ of a net $(x_i){i\in I}$ must have the following cofinality property. For all $i\in I$, there exists $j_0\in J$ such that $f(j)\geq i$ for all $j\geq j_0$. It is not enough to merely have $f(j_0)\geq i$ for a single $j_0$, which is how the wikipedia article defines subnets. With my definition, you can show that a cluster point is the same as a subnet limit. – Ehsaan May 03 '20 at 00:49
  • @Ehsaan You can have a look at these posts: Limit superior is a cluster point of a net, Cluster point of a net is a limit of a subnet. In needed, we can discuss this further also in the general topology chatroom. – Martin Sleziak May 03 '20 at 08:29
  • I feel that definition 1. by the OP may be not good because the collection of all subnets of a given net may not be a set but a class. Could you confirm if my understanding is correct? – Analyst Feb 23 '22 at 23:39
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    @Analyst What you say in your comment is true, but this can be easily solved - if we take limits of all convergent subnets, we definitely get a set. (It is a subset of $\mathbb R$.) And if we take supremum of all of those limits, this will be the limit superior. (So this is an equivalent definition of $\limsup$.) In the other word, to get $\limsup$, we take supremum/maximum of the set of all cluster points. In a similar way it works for filters. – Martin Sleziak Feb 24 '22 at 15:22