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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 directed 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 sets or $\mathcal C=$ linearly ordered sets, 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

1 Answers1

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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!

added
This came up again, so let me amplify my remark about non-Hausdorff spaces, and avoiding the axiom of choice (AC).

Let $(X, \mathscr X)$ and$(Y, \mathscr Y)$ be topological spaces, let $f : X \to Y$ be a function, let $a \in X$. For $a \in X$ let $\mathscr X_a = \{U \in \mathscr X : a \in U\}$ be the set of open neighborhoods of $a$; similarly for $\mathscr Y_b$. We say that $f$ is continuous at $a$ iff:

for every $V \in \mathscr Y_{f(a)}$ there exists $U \in \mathscr X_a$ such that $f(U) \subseteq V$.

But if we like to use nets, we want to say: this is equivalent to

for every net $(p_d)_{d \in D}$ in $X$, if $p_d \to a$ then $f(p_d) \to f(a)$.

To avoid AC we do it like this: Suppose $f : X \to Y$ is discontinuous at $a$. We want to find a net with $p_d \to a$ but $f(p_d) \not\to f(a)$. Easy: Let $$ D = \big\{(U,x) : x \in U, x \in U \in \mathscr X_a\big\} \tag1$$ ordered by $$ (U_1,x_1) \le (U_2, x_2) \quad\Longleftrightarrow\quad U_1 \supseteq U_2. \tag2$$ This is directed because: the intersection of two open sets is open. [We avoided AC here; we did not choose an element of $U$, instead we use all of the elements.]

Define a net $(p_d)_{d \in D}$ by $p_{(U,x)} = x$. Then we can show $x_d \to a$. Indeed, given $U \in \mathscr X_a$, we have $p_d \in U$ for all $d \ge (U,a)$.

Also we claim $f(p_d) \not\to f(a)$. We assumed $f$ is discontinuous at $a$. So there is $V$ with $f(a) \in V \in \mathscr Y_{f(a)}$ but for any $U \in \mathscr X_a$ we have $f(U) \not\subseteq V$. That is: there is $(U,y) \in D$ with $f(p_{(U,y)}) \notin V$.


Remarks. Notice that, in a non-Hausdorff space, it could be that there is a minimal open set with more than one element. The construction above still works. But the directed set ($1$) ordered by ($2$) is not partially ordered. Only quasi-ordered.

The originating material on nets (Birkhoff 1950, then Kelley 1955) used quasi-ordered directed sets, so everything worked. Subsequent readers of their expositions have sometimes not noticed this.

GEdgar
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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
  • So what would be an example of a cycle involving partitions with a Riemann integral? – The_Sympathizer Apr 14 '17 at 08:00
  • @mike4ty4 ... One partition, but two different choices of tags. Each is $\le$ the other in our ordering. but we don't care that they are $\ne$. – GEdgar Apr 14 '17 at 11:55
  • Really like your answer @GEdgar. Thinking about Riemann sums ... seems to me we could massage the structure of the tagged paritions of an interval back into a partially ordered directed set by defining < to mean strict refinement and = to mean "exactly the same" (including the tagging). (i.e. two different taggings of the same partition will not be comparable.) With these meanings, as I mentioned above, I think $\leq$ will give the collection of tagged partitions the structure of a partially ordered set with a direction. – Selrach Dunbar Jun 20 '19 at 23:14
  • @AsafKaragila That can be done in ZFC. Do you mean something specific with "space"? Like metric space? – Jakobian Dec 07 '23 at 18:38
  • @Jakobian: Yes. Metric space. Or, you know, $\Bbb R$. – Asaf Karagila Dec 07 '23 at 20:00