We recall at this point that every topology is a lattice, and indeed a complete Heyting algebra (remember O-3.22!). It is therefore meaningful to search for prime and co-prime elements in σ(L) (see 1-3.11, -3.15). To formulate one of our conditions it is necessary to speak of the continuity of an operation (the main topic of the next section). We say that the sup operation is jointly continuous with respect to the Scott toplogy provided that the mapping

(x, y) ↦ x ∨ y : (L, σ(L)) × (L, σ(L)) → (L, σ(L))

is continuous in the product toplogy.

1.11. Proposition. Let L be a complete lattice and U ⊆ L.

(i) U is a co-prime in σ(L) iff U ∈ σ(L) and U is a filter;

(ii) U is a prime in σ(L) and U ≠ L if U = L \ ↓u for some u ∈ L;

(iii) This last condition is also necessary, provided that the sup operation is jointly continuous; (iv) This is the case if L is a continuous lattice.

Proof. (i): Firstly suppose that U ∈ σ(L) is a filter and that U is not a co-prime in σ(L). Thus there are V, W ∈ σ(L) such that U ⊆ V ∪ W and elements v ∈ U \ V and w ∈ U \ W. Since V and W are upper sets we have vw ∉ V ∪ W. But vw ∈ U since U is a filter. This is a contradiction.

Secondly, suppose that U is a co-prime in σ(L). To show that U is a filter, note first that it is an upper set. Suppose v, w ∈ U. Then U ⊈ L \ ↓v and U ⊈ L \ ↓w. By 1.4(ii), the sets L \ ↓v and L \ ↓w are Scott open. Thus, since U is co-prime,

U ⊈ ((L \ ↓v) ∪ (L \ ↓w)) = L \ (↓v ∩ ↓w) = L \ ↓vw.

Thus U ∩ ↓vw ≠ ∅. Since U is an upper set, this means vw ∈ U.

(ii): Recall that L \ ↓u = L \ {u}^{−} for all u by 1.4(ii). But the complements of singleton closures are prime in any topology.

(iii): Assume that the sup-operation is jointly continuous relative to the Scott topology, and let U ∈ PRIME σ(L), U ≠ L. Set A = L \ U and u = sup A. In order to show that U = L \ ↓u it suffices to verify u ∈ A. Since A is closed under directed sups by 1.4(i), all we have to verify is that A is directed.

By way of contradiction assume that it is not. Then there are elements a, b ∈ A with a ∨ b ∈ U. By the continuity of the sup-operation we would find Scott-open neighborhoods V and W of a and b, respectively, such that V ∨ W ⊆ U. But since V and W are upper sets, we have V ∨ W = V ∩ W. Since U is prime, the relation V ∩ W ⊆ U implies V ⊆ U or W ⊆ U. This would entail a ∈ U or b ∈ U, which would contradict a, b ∈ A = L \ U.

(iv): Finally suppose that L is continuous. In order to show the continuity of the sup-operation at (a, b) we pick some u ≪ a ∨ b. By I-1.6 we have

a ∨ b = (sup ↓ ↓a) ∨ (sup ↓ ↓b) = sup(↓ ↓a ∨ ↓ ↓b)

Since ↓ ↓a ∨ ↓ ↓b is directed (I-1.2(iii)), we find some x ≪ a and y ≪ b with u ≪ x ∨ y (by I-1.19). But then ↑ ↑x and ↑ ↑y are Scott-open neighborhoods of a and b, respectively, such that

↑ ↑x ∨ ↑ ↑y ⊆ ↑(x ∨ y) ⊆ ↑ ↑u.

Since the ↑ ↑u with u ≪ a ∨ b form a basis of σ(L)-neighborhoods of a ∨ b by Proposition 1.10(i), the desired continuity is established. ⊓ ⊔

We can immediately rephrase 1.11 in topological terminology, if we recall the concept of a sober space (see I-3.36 and the discussion). Remember that a non-empty subset A of a topological space X is called irreducible iff it is closed and not the union of two proper closed subsets (that is, the complementary set X \ A ∈ PRIME O(X); see I-3.11). A space X is called sober iff every irreducible set A has a unique dense point (that is, A = {a}^{−} with a unique a ∈ A). Clearly all singleton closures are irreducible. (Notice that a sober space is automatically T_{0} since {x}^{−} = {y}^{−} always implies x = y.) We now have the following corollaries of 1.11 with a slight sharpening:

1.12. Corollary. If L is a complete lattice such that the sup-operation is jointly Scott-continuous, then (L, σ(L)) is a sober space.

Proof. Immediate from 1.11 and the definitions. ⊓ ⊔

1.13. Corollary. If L is a continuous lattice, then (L, σ(L)) is a quasicompact and locally quasicompact sober space. In particular, (L, σ(L)) is a Baire space.

Proof. We have to show that a point x ∈ L has a basis of quasicompact neighborhoods. By 1.10 the sets ↑ ↑y with y ≪ x form a basis for the neighborhoods of the point. But as we know, if x ∈ U ∈ σ(L), then actually we have a y ∈ U with y ≪ x; hence, ↑y ⊆ U, and so the sets ↑y can be used as neighborhoods. Since ↑y (and hence, in particular L = ↑0) is trivially quasicompact with respect to any topology whose open sets are upper sets, the assertion is proved. That (L, σ(L)) is a Baire space follows from I-3.43.10. ⊓ ⊔

We wish to warn the reader about a subtlety concerning the joint continuity of the sup-operation above. We cannot be satisfied by saying that the sup-operation is a continuous function (L×L, σ(L×L)) → (L, σ(L)); this continuity is weaker, since in general we have a proper containment of topologies: σ(L×L) ⊃ σ(L)×σ(L). We will return to this question at greater length in Section 4 below (see 4.11 ff).

We know enough about the Scott topology now to use it for yet another characterization theorem for continuous lattices.

1.14. Theorem. For any complete lattice, the following conditions are equivalent:

(1) L is continuous;

(2) If U ∈ σ(L), then U = ⋃{↑ ↑x : x ∈ U};

(3) Each point has a neighborhood basis (in the Scott topology) of Scott-open filters, and σ(L) is a continuous lattice;

(4) For each point x ∈ L we have x = sup{inf U : x ∈ U ∈ σ(L)}

(5) σ(L) has enough co-primes and is a continuous lattice;

(6) σ(L) is completely distributive;

(7) Both σ(L) and σ(L)^{op} are continuous.

Proof. (1) implies (2): Use 1.10.

(2) implies (1): Let x ∈ L and set y = sup ↓ ↓x ≤ x. If y < x, then L \ ↓y is a Scott-open neighborhood of x; hence by (2) it contains an open neighborhood ↑ ↑z of x with z ∈ L \ ↓y. But then z ≪ x, and thus z ≤ sup ↓ ↓x = y, a contradiction.

(1) implies (3): By (1) iff (2), x has arbitrarily small neighborhoods of the form ↑ ↑y with y ≪ x. By 1.10 and I-3.3, we know then that x has arbitrarily small Scott-open neighborhoods which are filters. In order to prove the continuity of σ(L), we let U be Scott open. For any x ∈ U we find a y ∈ U with y ≪ x by (2). Then x ∈ ↑ ↑y ∈ σ(L), and we claim that ↑ ↑y ≪ U: Indeed, if D is a directed family of Scott-open sets covering U, then one of its members must contain y, hence ↑y, since Scott-open sets are upper sets, and thus it contains ↑ ↑y. We have shown U = ⋃{V : V ≪ U}.

(3) implies (4): For x ∈ L set

y = sup{inf U : x ∈ U ∈ σ(L)} ≤ x.

If y < x, then L \ ↓y is a Scott-open neighborhood of x. Let V be a Scott-open neighborhood of x with V ≪ L\↓y, which exists since σ(L) is continuous by (3). Now use (3) to find a Scott-open filter neighborhood U of x within V. By the definition of y we have inf U ≤ y. Then

L \ ↓y ⊆ L \ ↓ inf U = L \ ⋂{↓u : u ∈ U} = ⋃{L \ ↓u : u ∈ U}.

Since U is a filter, the L \ ↓u for u ∈ U form a directed family of Scott-open sets. Since V ≪ L \ ↓y, there must be a u ∈ U such that V ⊆ L \ ↓u and so u ∉ V. This is a contradiction to U ⊆ V. Thus x = y.

(4) implies (1): Clear since for every Scott-open neighborhood U of x one has inf U ≪ x.

(3) iff (5): Clear from 1.11.

(5) iff (6) iff (7): Consequence of I-3.15. ⊓ ⊔

Note that condition (2) is stronger than saying that σ(L) has a basis of sets ↑ ↑x. A parallel result to 1.14 for algebraic lattices reads as follows:

1.15. Corollary. For any complete lattice L the following conditions are equivalent:

(1) L is algebraic.

(2) The Scott topology has a basis of sets ↑k where k ∈ K(L).

(3) σ(L) is algebraic and has enough co-primes.

Proof. (1) implies (2): Let U be a Scott-open neighborhood of x. We recall that x = sup(↓x ∩ K(L)) and ↓x ∩ K(L) is deirected. Hence by 1.2(ii), we find a k ∈ ↓x ∩ K(L) ∩ U. Then ↑k = ↑ ↑k is a Scott-open neighborhood of k, hence of x, with ↑k ⊆ U.

(2) implies (3): If k ∈ K(L), then ↑k ∈ σ(L) since ↑ ↑k = ↑k. Now ↑k is a quasicompact set (if ↑k is covered by Scott-open sets, then one of them must contain k, hence ↑k by 1.2(i)). Thus σ(L) is algebraic by I-4.4. Since all ↑k are filters, hence co-primes by 1.11, we are done.

(3) implies (1): Since σ(L) is algebraic, the Scott topology has a basis of quasicompact sets U. Since there are enough co-primes, U is a union of open filters by 1.11, and thus, by quasicompactness, U = U_{1} ∪ … ∪ U_{n} with open filters U_{k}. It is no loss of generality to assume that none of the U_{k} is contained in the union of the others. Then V = U_{1} \ (U_{2} ∪ … ∪ U_{n}) is quasicompact and filtered. We claim that V has a smallest element u_{1}.

For if not, then V ⊆ ⋃{L \ ↓v : v ∈ V }; and by quasicompactness and the fact that the L \ ↓v form a directed family, there would be a v ∈ V with V ⊆ L \ ↓v, notably v ∉ V, which is impossible. Since U_{2} ∪ … ∪ U_{n} is an upper set, it cannot contain inf U_{1}. Hence u_{1} = min U_{1}. Since U_{1} is an upper set, U_{1} = ↑u_{1}. Since U_{1} is Scott open, u_{1} ∈ K(L).

We have shown that σ(L) has a basis of sets ↑k where k ∈ K(L). Now let x ∈ L, set y = sup(↓x∩K(L)) ≤ x. If y < x, then the Scott-open neighborhood L \ ↓y of x would contain a basic neighborhood ↑k of x. Then there would be a k ∈ K(L) with k ≤ x and k ≰ y which contradicts the definition of y. ⊓ ⊔