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We know that in general topological spaces, topological concepts like closeness, compactness, continuity ,etc.. do not have an equivalent sequential description. However if we consider the generalized version of sequences which is called "Net" see https://en.wikipedia.org/wiki/Net_(mathematics), then everything become fine.

However working with net is not as convenient as sequences!! I have several questions regarding nets:

1-In definition of net in topological spaces, can we assume without loss of generality that the index set is totally ordered set (a chain)?

2- Let X be a metric space (you may assume X=R) and $\{x_n \}$ be a convergent sequence in X, what is a subnet of $x_n$ whose index set is an uncountable set?

3- Let $[0,1] \subseteq R$ of course there exists a net, say $\{x_i\}_{i \in I}$ such that $\{x_i\}_{i \in I} =[0,1] $, Could we choose this $\{x_i\}$ such that it is convergent ?

4-1- Let X be a metric space and $\{x_i\}_{i \in I}$ be a convergent net in $X$, can we derive a subnet of $\{x_i\}_{i \in I}$, say $\{x_m\}_{m=1}^{\infty}$ which itself is a sequence?

5- $\{x_i\}_{i \in I}$ be a convergent net in $X=R$ can we extract a subnet of $\{x_i\}_{i \in I}$ which is monotone is $R$?

6- $\{x_i\}_{i \in I}$ be a convergent net in the topological space $X$ can we extract a subnet of $\{x_i\}_{i \in I}$ which its index' set is totally ordered set?

Among all above questions, number $5,6$ is the most important one.

Thanks in advance for your nice help and answer .

Red shoes
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  • (1) No. (2) A sequence is a subnet of itself. (3) ? (4) Yes. – egreg Apr 29 '17 at 22:58
  • @egreg can you give an example to reject 1?

    In (2) actually wanted to say an uncountable index set, not infinite !

    I am almost sure that 4 is not true in general, you said yes can you prove it?

     – Red shoes Apr 29 '17 at 23:06
  • (3) are you asking if there exists a convergent subnet? Of course. Just take $x_i = .5$ for each $i \in I$. – mathworker21 Apr 29 '17 at 23:12
  • @mathworker21 I meant the net $ x_i = [0,1] $ itself ! which its range covers whole interval $[0,1]$ not just one point of it like $.5$ ! – Red shoes Apr 29 '17 at 23:21

1 Answers1

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  1. I'm not sure what you mean that "we can assume without loss of generality". Nets over non-linearly ordered sets occur naturally and dispensing of them would be ill-advised. If you are asking whether every net has a subnet ordered by a totally ordered set, then again, the answer is no. For example, if $\mathcal U$ is a non-principal ultrafilter of natural numbers, then $(\min A)_{A\in \mathcal U}$ is a net convergent to $\mathcal U$ in $\beta{\bf N}$, but a net of natural numbers ordered by a totally ordered set will have to revisit a number on a cofinal set, or it will have an increasing sequence as a subnet. In particular, it could not converge to anything but a natural number.
  2. By definition, a subnet of $(x_n)_{n\in {\bf N}}$ is a net $(x_i')_{i\in I}$ such that there is a function $f\colon I\to {\bf N}$ such that $f(i)\leq f(j)$ whenever $i\leq j$ and $f(i)\to \infty$ as $i\to \infty$. What is it you are confused about?
  3. Yes, for example you can take $I=[0,1]$ and $x_i=i$.
  4. No. For example, if $I=\omega_1$ and $x_i$ is a constant net, then trivially $(x_i)_i$ is convergent, but because cofinality of $\omega_1$ is not $\omega$, there is no subnet which is a sequence.
  5. Convergence of the net is not important here. This is true, by essentially the same argument as the one used in this answer. Either there is a cofinal set of $i$ such that for all $j\geq i$ we have $x_j\geq x_i$, or not. If yes, we are done. If not, for all sufficiently large $i$, we have $j>i$ with $x_j<x_i$, and we can recursively define a decreasing subnet.
tomasz
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  • by W.L.O.G, I meant, can we define nets with this extra assumption that is the index set are totally ordered, and we are still able to characterize closeness, compactness, and continuity using this type of net ? If no, Is there any specific category of topological spaces that their topological concepts can be described equivalently by these type of nest? – Red shoes Apr 30 '17 at 00:00
  • for 2, I want an example? – Red shoes Apr 30 '17 at 00:02
  • @nonlinearthought: I doubt it in general, though I don't have a counterexample offhand (a slight modification of the one I gave should work, though -- for example, you could double a single ultrafilter in $\beta{\bf N}$ to obtain a non-Hausdorff space, but since all convergent "totally ordered" nets are eventually constant...). I suppose for first-countable spaces that is enough, but I doubt there is any larger interesting category in which you could do that. For 2, you can take $I={\bf R}+$, $x'_i=x{\lfloor i\rfloor}$. – tomasz Apr 30 '17 at 00:12
  • ,thanks any way, what do you think about 6? – Red shoes Apr 30 '17 at 00:28
  • @nonlinearthought: You really should not keep modifying your question like that. Besides, I already gave you a counterexample for 6. – tomasz Apr 30 '17 at 18:24
  • I didn't modify anything !!! maybe few typos, I only added part 6 to the question, since didn't want post new question in website this is not called modifying! BTW if I modified things your answer wouldn't work for this problem. – Red shoes Apr 30 '17 at 18:41
  • @nonlinearthought: There's nothing wrong with asking another question. – tomasz Apr 30 '17 at 18:48
  • we both don't know if adding a part to a question is wrong but asking Numerous same type of questions is OK ! One thing is clear here is that if same type of questions are gathered in one place it is more convenient for people to take advantage of answers of similar questions they may have! That's why mathstack suggests first to search in website your questions then if you couldn't find it post new question to avoid of overwhelming here by millions similar questions. – Red shoes Apr 30 '17 at 19:18
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    @nonlinearthought: I'm pretty sure adding more parts to an existing question is generally frowned upon (apart from things like clarifications etc.). In fact, asking multiple loosely related questions in one thread is already somewhat frowned upon. If you have another, related question, you should just add links both ways. – tomasz Apr 30 '17 at 19:23