last day (19 days later) » 

8:48 AM
room topic changed to Limsups of nets: Question math.stackexchange.com/questions/188722/limsups-of-nets/188738 [general-topology] [limsup]
2
A: Limsups of nets

Martin SleziakAs 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 clust...

Instruction about how to read TeX in chat can be found here.
Jeff wrote in his comment
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 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?
Let me talk about directed sets with maximum elements and the nets for a while.
*******
So let us assume that we have a net $(x_d)_{d\in D}$ and that $m$ is the maximal element of $D$.
I claim that:
A. $x_m$ is the limit of this net
B. $x_m$ is the only cluster point.
All nets we are working with are nets of real numbers.
Let us have a look at the exact definition of cluster points.
A real number $c$ is a cluster point of $(x_d)$ if for each neighborhood $U$ of $c$ and for each $d_0\in D$ there exists $d\ge d_0$ such that $x_d\in U$.
So it is very easy to see that $x_m$ is a cluster point, since $x_m$ belongs to each neighborhood of $x_m$ and we have $m\ge d_0$ for each $d_0\in D$.
On the other hand, suppose that $c$ is a cluster point of the net $(x_d)$. What does this exactly means?
Let us have a look what the definition says for $m=d_0$. We get that every neighborhood of $c$ must contain $x_m$.
And this implies that $c=x_m$.
*************
Showing that $\lim_{d\in D} x_d=x_m$ is very similar.
And $\limsup_{d\in D} x_d=x_m$ should be relatively easy, too.
************************
I'm leaving for lunch now, but feel free to ping me if a more detailed explanation of something is needed. (Or if I made a mistake somewhere.)
 
9:34 AM
**********************
BTW I should have probably also mentioned that every subnet will contain the element $x_m$. (Every cofinal subset of $D$ contains $m$.)
I am not sure what is meant by this part of your comment: I am still surprised that the subnet limit definition is not equal to the sup of the tail ends.
Since I claim in my answer that we have for every net $$\limsup x_d = \lim_{d\in D} \sup_{e\ge d} x_e = \inf_{d\in D} \sup_{e\ge d} x_e.$$
 
 
7 hours later…
4:27 PM
@MartinSleziak ah okay I understand what you mean now. I thought by cluster point you meant limit point or accumulation point. If someone said those things then what would most likely be meant is limit point of the image of the net. Okay, now everything you're saying is making a lot of sense, particularly why the cluster point limsup is equivalent to the other two for sequences. Thanks. Oh by the way, I hope chats never disappear since I keep records of what I learn.
 
 
1 hour later…
5:31 PM
@Jeff According to this Rooms will exist indefinitely, so long as there is at least one person actively talking in the room. A room is considered worth retaining if it has more than 15 messages by at least 2 users.
I'd say we're fine past that.
 

  last day (19 days later) »