Totally-Ordered Nets
August 3rd, 2007 by WaltI was thinking about nets the other day, when I was reminded of something that I wondered when I first encountered them. Is the generalization to partially-ordered sets strictly necessary? The generalization to partially-ordered sets is useful, but are there spaces with points that are not reachable by totally-ordered nets alone?
In a first countable space, a point lies in the closure of a set if and only if there is a sequence of points in the set that converges to the point. This property fails in general. For example, an uncountable set with the cofinite topology has no non-trivial convergent sequences, but the closure of any infinite set is the whole space. You can recover this property if you pass to nets, which allow fairly general partially-ordered sets to be the index set (the only requirement you must impose is that they be directed sets). So if you require the index sets of your nets to be totally ordered, is there a space which contains a point that is not the limit of such a net?
Poking around Wikipedia, I found that the page for order topology, which suggests that the Tychonoff plank is an example of a space where totally-ordered nets are not sufficient. The page discusses nets indexed by ordinals, which possibly is a loophole, but it seems like a very narrow one. I’d be curious if anyone knows for sure.
August 3rd, 2007 at 2:11 am
Hmm, I wonder. Given any net, consider its restriction to a maximal chain (totally ordered set) within the index set. Your question seems to boil down to the question of whether a such a maximal chain “reaches the end” of the index set (or can be chosen to do so). In other words, given an element of the net’s index set, is there an element of the chain that is at least as large? I haven’t thought through the details, but I guess the answer to this question will settle your question either way.
August 3rd, 2007 at 2:40 am
I remember wondering about this too, when I first encountered nets. I thought about it a lot… but that was decades ago, and I forget the upshot.
Right now I’m thinking there should be spaces where non-totally-ordered nets are necessary to fully describe the topology. Following Harald Hanche-Olsen’s lead, I think we should try to construct a directed directed set in which no chain is cofinal.
Perhaps the Tychonoff Plank will do the job.
August 3rd, 2007 at 4:24 am
Oh, wait a minute. The Tychonov plank might be a good example after all. Consider the directed set ω×ω1 with the ordering in which (a,b)<(c,d) iff a<c and b<d: Any chain must be countable, so there is a definite upper limit on the second component of the members of the chain. In other words, it doesn’t “reach the top”. So this settles my question in the negative, but does it settle yours?
I think so, for if you consider any net with a totally ordered index set, then you can find an equivalent net indexed with ordinals. Just wellorder the index set, and inductively keep or throw out members, keeping just those who are greater (in the original ordering) than any element previously kept. The result is a well-ordered subset of the original index set, which reaches “all the way up” in the original. Right?
Uh, no preview on your blog. I’ll just have to toss this in and see if my HTML formatting is accepted or what.
August 3rd, 2007 at 7:49 am
In omega_1 with cofinite topology, the integers converge to omega. So in the example for “sequences are not enough”, you mean an uncountable set with cocountable topology, right? Then you can isolate the would-be limit from all members of the sequence. Then the space doesn’t have non-trivial convergent sequences; but every uncoubtable set is dense.
(My favourite example for this is [0,omega_1] in the usual order topology, where you can see right away that omega_1
is in the closure, yet no sequence can converge to it.)
As for our question: not only are transfinite sequences (indexed by ordinals) not sufficient - even if you allow the
sequence to be indexed by any linear order, it is not sufficient. The same example (Tychonoff plank) works, with
the proof trivially modified as follows.
The point p = (omega1, omega) of the Tychonoff plank is in the closure of the open set U = omega1 x omega. Let’s suppose that some sequence (a_x, n_x) of points from U, indexed by members x in L of some linear order L, converges to p. Surely L needs to be uncountable; in fact, the cofinality of L must be uncountable.
L decomposes into countably many sets L_n = {x | n_x = n }. One of these sets must be unbounded in L: if not,
take s_n to be an upper bound for L_n in L. Then {s_n | n in omega } is a countable set cofinal in L.
So take the unbounded L_n. The sequence of a_x for x in L_n surely converges to omega1. So the sequence
whose members are the (a_x, n_x) = (a_x, n) for x in L_n, which is a subsequence of the original sequence,
converges to (w1, n). Contradiction.
The only place where we changed the original proof (that no transfinite ordinal sequence in U can converge
to p) is that we are taking upper bounds of L_n in the linear order L, instead of suprema in the ordinal.
August 3rd, 2007 at 10:20 am
Off the top of my head (so quite probably wrong): why not take the Stone-Cech compactification of a countable discrete set (say the natural numbers N for sake of definiteness)?
My *guess* is that there must be ultrafilters on N which are not limit points of sequences of principal ultrafilters — tho’ I haven’t thought about this properly.
August 15th, 2007 at 8:21 am
As Harald’s example shows, a directed set need not contain a cofinal subset indexed by a chain. However, every directed set contains a monotonically cofinal subset indexed by a tower.
A tower is a poset (T,
August 15th, 2007 at 8:23 am
[excuse my stupid use of < in HTML]
As Harald’s example shows, a directed set need not contain a cofinal subset indexed by a chain. However, every directed set contains a monotonically cofinal subset indexed by a tower.
A tower is a poset (T,≤) that is isomorphic (as a poset) to
the set of all finite subsets of some suitable set (ordered by inclusion).
Given any directed (X,≤), the set T of all finite subsets of X itself
is surely a tower. Now let’s create a cofinal subset {x_t | t in T} in X
that is even _monotonically_ cofinal, that is,
[M] for s a (strict) subset of t, x_t ≥ x_s in X.
This can be done by induction on the size of t: for |t| = 1, that is,
t = {x} for some x in X, take x_t = {x}. This satisfies [M].
Once you have x_t for all |t| = n, define x_t for |t| = n + 1 as follows:
consider all (strict) subsets s of t, and the (monotonically) corresponding
x_s. Then take x_t in to be greater than all of these x_s (which can be done,
because X is directed). In this fashion, you create the {x_t | t in T}.
This satisfies [M] by definition. Also, it is surely cofinal.
This appears as an exercise in Birkhoff’s Lattice Theory
(crediting M. Krasner)