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)

A tower is a poset (T,

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.

(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.

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.

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.