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Question: Usually net is defined as a function from a directed preordered set to a topological space. What would we lose or gain if we worked with partially ordered directed sets only?


Background and motivation

For me, one of the most important property of nets in topological spaces is the following:

Let $X$ be a topological space, $x\in X$ and $A\subseteq x$. Then $x\in\overline A$ if and only if there exists a net $(x_d)_{d\in D}$ such that $x_d$ converges to $x$ and $x_d\in A$ for each $d\in D$.

The above property means that topology of $X$ is uniquely determined by convergence of nets.

The above property would be true also if we worked only with nets on preordered posets.

Indeed, let $x\in\overline A$. Let $\mathcal N_x$ be the set of all open neighborhoods of $x$. Then $D=(\mathcal N_x,\supseteq)$ is a poset, which is directed. For any $U$ choose one element $x_U\in A\cap U$. Then the net $(x_U)_{U\in\mathcal N_x}$ converges to $x$.

So, if I did not make a mistake in this proof, the above property is true also if we work only with nets on posets.


In some constructions it is useful that we are allowed to work with nets on preordered sets. One of them is constructing a net corresponding to a given filter $\mathcal F$ such that this net converges to $x$ if and only if the filter $\mathcal F$ does. Here the net on the set $\{(A,a); a\in A\in \mathcal F\}$ preordered by $$(A,a)\le (B,b) \Leftrightarrow A\supseteq B$$ is used, and this is indeed a preordered set, not a partially ordered set. See the part on correspondence of nets and filters in P. L. Clark's notes on convergence for more details and other related constructions. In fact, P. L. Clark writes here that this "justifies our willingness to entertain directed quasi-ordered sets". (This answers to some extent my question. But I guess that the authors working with posets also have some reasons to do so.)


Another possible viewpoint:

For any class $\mathcal C$ of directed sets we can define the class of topological spaces defined by the property:

$V$ is closed in $X$ if and only if $V$ is closed w.r.t. limits of nets on directed set from $\mathcal C$.

  • If $\mathcal C$ = all directed sets, then we get all topological spaces.

  • If $\mathcal C=\{\mathbb N\}$, then we get sequential spaces

  • If $\mathcal C=$ well-ordered set or $\mathcal C$=linearly ordered set, then we get pseudoradial spaces

So in this context we can view this question as the question what we get if $\mathcal C$=directed posets (which seems quite natural, after seeing that several classes of spaces arise in this way).


Going through literature and also quick google searches bellow shows that quite a few authors use posets and not preordered sets.

I could mention a few books which use posets:

  • Munkres: Topology, Supplementary Exercises after Chapter 3, p. 187 in 2nd edition. (Although he does not pay much attention to nets - they are left as supplementary exercises.)

  • Runde: A taste of topology, see Definition 1.3.3 on p.18 and Definition 3.2.8 on p.73

  • Bukovský: The Structure of the Real Line p.32

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Somewhat related –  t.b. Nov 23 '11 at 17:45

1 Answer 1

In Hausdorff spaces, we probably lose little either way. In the "partial order" setting we require the axiom of choice more often. (Did you see "choose" in your write-up? Now do the same thing WITHOUT the axiom of choice...)

There is more than one paper where the author claims something doesn't work for nets (in order to justify his/her use of filters) by incorrectly requiring the partial-order version of directed set.

One use of nets (indeed, the original use by Moore and Smith) is to explain the convergence used in the definition of the Riemann integral, where you have a Riemann sum for each tagged partition of an interval, and you direct them by refinement, allowing any choice of tags. Preordered only!

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Yes, I've noticed that I was using AC, but thanks for pointing it out explicitly. I do not know, whether the same is possible in ZF - in fact I did not think about it; but if I'm working in ZFC this should not be much of a hindrance. (Topology without AC is quite weird anyway.) –  Martin Sleziak Nov 23 '11 at 17:57
    
@Martin: Quite is a big understatement... :-) (Trivial example: Separable space with a subspace which is not separable.) –  Asaf Karagila Nov 23 '11 at 18:00

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