]> All finite distributive lattices occur as intervals between Hausdorff topologies

All finite distributive lattices occur as intervals between Hausdorff topologies

R. W. Knight, P. Gartside and D. W. McIntyre

October 1997
Abstract

It is shown that a finite lattice L is isomorphic to the interval between two Hausdorff topologies on some set if and only if L is distributive. The corresponding results had previously been shown in ZFC for intervals between T1 topologies and, assuming the existence of infinitely many measurable cardinals, for intervals between T3 topologies.

1 Introduction

The collection Σ(X) of all topologies on a given set X forms a lattice under inclusion, in which the meet of two topologies is their intersection, while the join is the topology with their union as a sub-basis. This lattice has been an object of study since it was first defined by Birkhoff [1]: see the survey article by Larsen and Andima, [4], for more details.

In this paper we are concerned with the local structure of this lattice. Specifically, we are looking at the problem of determining the finite lattices L such that there exist some set X and some σ, τ Σ(X) (perhaps with some specific separation properties) such that the interval [σ, τ] = { μ Σ(X) | σ μ τ } is isomorphic to L.

This problem was solved for T1 topologies σ and τ by Valent and Larson and Rosický: Valent and Larson [7] proved that any finite distributive lattice can be realized as an interval between T 1 topologies, and Rosický [5] proved that any finite interval between T 1 topologies must be distributive. Hence a finite lattice can be realized as such an interval if and only if it is distributive.

Of course, Rosický's result implies that every finite interval between Hausdorff topologies must be distributive. In this paper we will show that the converse is true. This result has already been shown by Good, McIntyre and Watson in [3]—indeed, the topologies σ and τ can be assumed to be T3—under the assumption of the existence of infinitely many measurable cardinals. In this paper we will make no set-theoretic assumptions beyond ZFC.

2 Notation, terminology and basic results

For P a partially ordered set, let O(P) denote the set of down-closed subsets of P, partially ordered by inclusion. For L a finite lattice let J(L) denote the set of join-irreducible elements (i.e. elements a such that a is not the least element of the lattice and, if a = b c then a = b or a = c). Recall that a finite lattice L is distributive if and only if LO(P) for some P, which happens if and only if LO(J(L)).

If F is a family of subsets of a set X with the finite intersection property, let F denote the filter on X which has F as a filter subbasis. We will say that A is compatible with F if F {A} has the finite intersection property.

Let u denote the least cardinal of a filter subbasis for a free ultrafilter on ω. It is easy to show that ω < u c. For more details, and relationships between u and other small cardinals, see [6]

If p and q are ultrafilters on sets A and B respectively, let p · q denote the ultrafilter { S A × B | { a A | Sa q } p }, where Sa = { b B | a, b S }. We define ultrafilters pn on An for n 1 by p1 = p, pn+1 = pn · p. Note that this product is associative: if p, q and r are ultrafilters, then p · (q · r) = (p · q) · r.

If X is a topological space, then an o-filter on X is a filter in the partial order of non-empty open sets of X, and an o-ultrafilter is a maximal o-filter.

For μ a topology on X and p X, let Nμ(p) denote the neighbourhood filter at p in the topology μ, and let Nμo(p) denote the o-filter of open neighbourhoods of p in the topology μ (in other words, Nμo(p) = Nμ(p) μ). For

F a filter on X and A X, let F A be the trace of F on A, in other words the family { F A | F F }. Notice that Nμ(p) A is a proper filter (i.e. does not contain ) if and only if p Aμ, and similarly for Nμo(p) A.

If σ is a topology on X and A X, let σ, A denote the topology which has σ {A} as a subbasis.

Good, McIntyre and Watson proved the following result in [3]:

Lemma 1 Let P be a finite partially ordered set. Suppose we can find a set X of the form X = {x0} aP Xa, where the sets Xa are disjoint, non-empty and do not contain x0, and a topology σ on X such that
  1. For each a P, Xaσ = {x 0} baXb.
  2. For each a P, Nσo(x0) Xa is an o-ultrafilter on Xa.

Let τ = σ, {x0}. Then [σ, τ]O(P).

Conversely, if O(P) can be realized as an interval between T1 (resp. T2, T 3) topologies, then it can be realized by a set X and T1 (resp. T2, T 3) topologies σ and τ of this form.

Suppose that P is a finite partially ordered set, and Q P. Let X = {p} aP Sa and σ, τ be as above. Put Y = {p} aQSa. Then SaσY = S aσ Y and Nσ Y o(p) Sa = Nσo(p) for a Q. Thus (Y, σ Y ) has the form required by Lemma 1 to ensure that [σ Y, τ Y ]O(Q).

Thus, to show that O(P) can be realized, it is enough to show that O(P) can be realized for some P containing P. We will show in the next section that, for every n ω, O(2n) can be realized as an interval between T2 topologies.

Definition 1 A nice basis for an ultrafilter p is a sequence Bα | α κ such that { Bα | α κ } is a filter basis for p and, for every α κ, Bα{ Bβ | β < α }.
Lemma 2 There exists an ultrafilter p on ω, a nice basis Bα | α u for p, and sets Sα,n for α u, n ω such that for each α u, { Sα,n | n ω } is a disjoint family of subsets of ω \ Bα and { Bβ | β < α } {Sα,n} has the finite intersection property for every n ω.
Proof. Let Cα | α u be a basis for a free ultrafilter p on ω of minimal cardinality. Let I = { α u | Cα{ Cβ | β < α } }. One can easily show that Cα | α I forms a nice basis for p, so I has cardinality (and hence order-type) u. Relabel the elements without changing their order as Bα | α u.

Now, for each α u, Bα{ Bβ | β < α }, so ω \ Bα is compatible with { Bβ | β < α }. However, { Bβ | β < α } {ω \ Bα} is too small to be an ultrafilter subbasis, so there exist disjoint sets A0 and B0 which are compatible with { Bβ | β < α } {ω \ Bα}. Put Sα,0 = (ω \ Bα) A0. Since { Bβ | β < α } {ω \ Bα, B0} is too small to be an ultrafilter subbasis, there exist disjoint sets A1 and B1 which are compatible with { Bβ | β < α } {ω \ Bα, B0}. Put Sα,1 = (ω \ Bα) B0 A1. Continuing in this way we construct a sequence Sα,n | n ω of disjoint subsets of ω \ Bα, each of which is compatible with { Bβ | β < α }, as required.  _

3 The main result

Throughout this section, A will denote some fixed finite set, and P = A, ordered by reverse inclusion. We will construct sets and ultrafilters as in Lemma 1 to show that O(P) can be realized as an interval between T2 topologies.

For each a P, let Xa = u|a| × ω × {a}. Let p, Bα | α < u and { Sα,n | α u, n ω } be as in Lemma 2. Let f : ω × P ω be a bijection. For each α u, n ω and a P let Sα,n,a = Sα,f(n,a) and let qα,n,a be a free ultrafilter extending {Bβ | β < α } {Sα,n,a}. Let r be a uniform ultrafilter on u (in other words, an ultrafilter extending the co-< u filter on u). Let x0 be some point which is not an element of aP Xa. For n ω and s un+1, let l(s) = s(n) (the “last” element of the sequence s), and let s ̄ = s \ {n, l(s)}. For a P \ {} let c(a) = { a \ {x} | x a } be the set of covers of a in P.

We will specify a topology σ on X = {x0} aP Xa by specifying a weak neighbourhood system for σ, in other words a collection B(x) of subsets of X containing x for each x X, such that a subset U of X is open if and only if for every x U there is some B B(x) with B U.

For x = , n, X, let B(x) = {{x}}.

For x = s, n, a Xa (where a P \ {}), let B(x) = { N(x, S) | S ql(s),n,a }, where N(x, S) = {x} { s̄, m, b | m S, b c(a) }.

Finally, let B(x0) = { N(x0, S) | S r|A| · p }, where N(x0, S) = {x0} { s, n, A | s, n S }.

It is easy to see that all the families B(x) are closed under finite intersections, so { B(x) | x X } forms a weak neighbourhood system for a topology σ on X. It is also clear that for each a P, Xaσ = {x 0} baXb.

Lemma 3 σ is a T2 topology on X.
Proof. For x = s, n, a Xa, let Ux = {x} { s |b|, m, b | m ω, a < b }. Then Ux is an open set containing x, and if y Ux then Uy Ux.

For each s un (where n |A|), let Y s = {s} × ω × Pn, where Pn = { a P | |a| = n }. We will prove by induction on n that Y s can be covered by a disjoint family { V x | x Y s } of open sets such that x V x Ux. The base step is trivial, since Y = X contains only isolated points. So suppose the result holds for all s un, and that s un+1. For each x = s, n, a Y s let V x = {x} { V s̄,m,b | m Sl(s),n,a, b c(a) }. Then x V x Ux, and V x is open. It remains only to show that these sets are disjoint. So let x = s, l, a and y = s, m, b be distinct elements of Y s, and suppose that z = t, k, c V x V y. Then we must have zx and zy, so for some i Sl(s),l,a and some d c(a), z V s ̄,i,d and similarly for some j Sl(s),m,b and some e c(b), z V s ̄,j,e. By inductive hypothesis, then, we must have s ̄, i, d = s ̄, j, e, so i Sl(s),l,a Sl(s),m,b. But these sets are disjoint, unless l, a = m, b, in which case x = y, a contradiction.

Now let x = s, m, a and y = t, n, b be distinct elements of X \ {x0}. Without loss of generality we assume that |a| |b|. We will consider separately two cases, depending on whether or not s = t |a|.

Case 1
s = t |a|. Let W1 = zY s\{x}V x, W2 = { t |c|, l, c | l ω, |a| < |c| |b| }, and W = W1 W2. Notice that if u, l, c W2 with |c| > |a| + 1 then {u ̄} × ω × c(c) W2, and if u, l, c W2 with |c| = |a| + 1 then {u ̄} × (ω \ {m}) × c(c) W1. Thus W is open, so x and y are separated by the disjoint open sets V x and W.
Case 2
st |a|. Let l be the least element of |a| such that s(l)t(l). Let α = s(l), β = t(l). If α < β, let Z0 = ω \ Bα and let W0 = Bα. If β < α let Z0 = Bβ and let W0 = ω \ Bβ. Either way, Z0 and W0 are disjoint subsets of ω, and for every s l + 1, k, c Ux Y sl+1, Z0 qα,k,c, while for every t l + 1, k, c Uy Y tl+1, W0 qβ,k,c. Put Z1 = { V sl,k,c | k Z0, c Pl }, and put W1 = { V tl,k,c | k W0, c Pl }. Put Z2 = { s |c|, k, c | l < |c| |a| } and W2 = { t |c|, k, c | l < |c| |b| }. Then, as in the previous case, Z = Z1 Z2 and W = W1 W2 are open sets. Moreover, since Z0 and W0 are disjoint, Z1 and W1 are disjoint, while since s(l)t(l), Z2 and W2 are disjoint. Thus Z and W are disjoint open sets separating x and y.

To complete the proof that σ is a T2 topology, we must show that we can separate x0 from any other element of X. So let y = s, n, a Xa. If a = then (since ω \ {n} in qα,m,b for all α, m and b, X \ {y} is open, so x0 and y are separated by the disjoint open sets X \ {y} and {y}. So suppose that |a| > 0. Let α = s(0). Let Z = {x0} { t, m, b | b P \ {}, t u|b|, t(0) > α } { , m, | m B α } and let W = (Uy \ Y ) { , m, | m ω \ Bα }. Then Z and W are disjoint open sets separating x0 and y.  _

Lemma 4 Nσ0(x0) Xa is an ultrafilter for each a P.
Proof. We will show that Nσo(x0) Xa = { T × {a} | T r|a| · p }. Since x0 Xaσ and r|a| · p is an ultrafilter, it is enough to show that, for each T r|a| · p, there is an open set U containing x0 with U Xa = T × {a}. Let l = |a|. We will construct a sequence Tl, Tl+1, , T|A| of sets such that
  1. Tl = T;
  2. if l i < |A| then for every s, n Ti+1 and every b Pi+1, { m ω | s ̄, m Ti } ql(s),n,b; and
  3. Ti ri · p for each i.

If we then take U = |b|<aXb i = l|A|Ti × Pi {x0} , then U will be the open set we require.

So, suppose we have chosen Tl, , Tk satisfying (1)-(3), and we wish to choose Tk+1. For s uk let (Tk)s = { n ω | s, n Tk }. Let S = { s uk | (Tk)s } p. Then, since Tk rk · p, S rk. For each s S choose αs u with Bαs (Tk)s. Let S = { s {k, β} | s S, α s < β < u }. Then, since { β | αs < β < u } is in r for each s S, s rk · r = rk+1. Put tk+1 = S × ω. Then { s uk+1 | (Tk+1)s p } = S rk+1, so Tk+1 rk+1 · p. Finally, suppose that s, n Tk+1 and b Pk+1. Then s S, so s ̄ S and αs ̄ < l(s). Thus Bαs ̄ ql(s),n,b and Bαs ̄ (Tk)s ̄, so { m ω | s ̄, m Tk } ql(s),n,b, as required.  _ Thus, putting τ = σ, {x0}, we have that [σ, τ] is isomorphic to O(P), as claimed. Now, since any finite partial order can be embedded in an order of the form A (ordered by reverse inclusion) for some finite set A, we have proved the following:

Theorem 1 Let L be a finite lattice. Then L is isomorphic to an interval between T 2 topologies if and only if L is distributive.
Proof. Necessity follows from Rosický's result that any finite interval between T 1 topologies is distributive. Sufficency follows from the previous remarks and the remark following Lemma 1  _

4 Conclusions and questions

The construction given in this paper shows that any finite distributive lattice can be realized as an interval between T 2 topologies on a set of size u. It clearly does not yield a regular topology: the closure of any neighbourhood of x0 will contain co-< u many elements of each Xa, not just r|a| · p many. Two obvious directions for possible improvements on this result are to ask for more separation on the topologies or to ask for smaller sets.

In [3] it is shown that the distributive lattice cannot be realized as an interval between T 3 topologies on a countable set. The question remains whether this lattice can be realized as an interval between T2 topologies on a countable set, and whether it can be realized as an interval between T 3 topologies on some set without assuming the existence of a measurable cardinal.

Acknowldegements

The work presented in this paper is one outcome of a mini workshop organised by the Mathematics Department of the University of Auckland in April 1996. The first author would like to acknowledge the financial assistance of the Mathematics Department of Auckland University, without which his attendance would not have been possible. The second author was present on a Royal Society Postdoctoral Fellowship. These authors wish to sincerely thank the Mathematics Department for their warm hospitality.

References

[1]   Birkhoff, G., On the combination of topologies, Fund. Math. 26 (1936), 156-166.

[2]   Comfort, W.W. & Negrepontis, S., The theory of ultrafilters, Springer-Verlag, Berlin, 1974.

[3]   Good, C., McIntyre, D.W. & Watson, W.S., Measurable cardinals and finite intervals between Hausdorff topologies, University of Auckland Mathematics Department Report Series 334, March 1996.

[4]   Larson R.E. & Andima S.J., The lattice of topologies: a survey, Rocky Mountain J. Math., 5 (1975), 177-198.

[5]   Rosický, J., Modular, distributive and simple intervals of the lattice of topologies, Arch. Math. Brno 11 (1975), 105-114.

[6]   Vaughan, J., Small uncountable cardinals and topology, in: G.M. Reed and J. van Mill, eds., Open Problems in Topology (North-Holland, Amsterdam, 1990) 195-218.

[7]   Valent, R & Larson, R.E., Basic intervals in the lattice of topologies, Duke Math. J. 39 (1972), 401-411.