The need for nets in topology

Question

I remember when I first heard about nets in topology (called also Moore-Smith sequences). I was told that most of useful topological properties which can be exressed in terms of sequences in the context of metric spaces can be generalized and expressed in the terms of nets: for example you can check continuity of mapping $f$ by checking whether $f(x_s) \to f(x)$ when $x_s \to x$ where $(x_s)_s$ is net. I was also warned about facts which no longer are true in the context of nets: for example if $x_s \to x$ then the set $\{x_s\}_s \cup \{x\}$ need not to be compact. I thought that this sort of things are due to the fact, that $(x_s)_s$ may be uncountable and the indexing set $S$ is only partially ordered instead of totally ordered. However even if you consider nets $(x_s)_s$ where $s$ runs over $\{0,1\} \times \mathbb{N} $ with lexycographic order (this is well ordered and countable set!) you can easily construct nets which are convergent and not bounded, with arbitary big closure etc. It raises the question: maybe we don't need to define nets as mappings $x:S \to X$ where $S$ is only partially ordered?

However in most of the proofs using nets, one usually uses the family of all possible neighborhoods of the given point (with reverse inclusion) as the indexing set which is partially ordered. Obviously in the context of metric spaces each point has a countable system of neigborhoods consisting of open balls with radius $\frac{1}{n}$ centered at that point. Therefore we can use standard sequences instead of general nets. Moreover, instead of writing $x_{B(x,\frac1n)}$ we simply write $x_n$. This system of neigborhoods has the property of being totally ordered.

Let us now consider the following example: $X=\beta \mathbb{N}$ the Stone-Cech compactification of the discrete countable space $\mathbb{N}$. This space is of cardinality $2^{\mathfrak{c}}$ but is also separable: $\mathbb{N}$ is the countable dense set in $X$. Therefore for each element $x \in X$ there is a subnet of $(n)_n$ which converges to $x$. Such a subnet is of the form $(n_U)_U$ where $U$ runs over the system of neigborhood of $x$. If for every $x$ the corresponding system of neighborhoods of $x$ could be totally ordered, we would get that $(n_U)_U$ is just $(n_k)_k$ (many terms are repeated) and thus there are only $\mathfrak{c}$ of possible subsequences of $(n)_n$ and therefore the closure of $\mathbb{N}$ cannot be of cardinality $2^{\mathfrak{c}}$. This contradiction shows that there are points $x$ such that their system of neighborhoods cannot be totally ordered.

EDIT: so to summarize things: we now from topology that $\overline{A}=\{ \lim_s a_s: (a_s)_s \ is \ a \ net \ in \ A \}$. So my question is:

1. Whether the following is correct: "If we define nets in the topological space $X$ as maps $x:S \to X$ where $S$ must be totally ordered, then the above formula for the closure does not hold and the counterexample is every separable topological space of cardinality $2^{\mathfrak{c}}$".

And also

2. Is it correct that in any such space $X$ (separable, of cardinality $2^{\mathfrak{c}}$) there must be points with the property that their bases of neighborhoods cannot be totally ordered?

I hope that now everything is clear. Sorry for the previous inprecise formulation.

Answer 1

There seems to be a gap in your argument where, under the assumption that the system of neighborhoods could be totally ordered, you seem to use that this total order must be countable. To fill the gap, note that any total order has a cofinal well-ordered subset and that the neighborhoods correspond to the sets in $U$. So you'd have a well-ordered (by reverse inclusion) family of subsets of $\mathbb N$, and such a family is necessarily countable.

Your argument could be simplified a bit, using the known fact that no nonprincipal ultrafilter has a countable base.

Note also that it is consistent (for example it follows form the continuum hypothesis) that some non-principal ultrafilter on $\mathbb N$ has a base totally ordered by inclusion modulo finite, i.e., by the relation $A\subseteq^*B$ that means $A-B$ is finite. It's only total ordering by genuine inclusion that's impossible for a nonprincipal ultrafilter base.