Nets Via Sequences in Topological spaces

Question

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 .

Answer 1

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.