On lattices of convex sets in ℝ^{n}

George M. Bergman

To the memory of Ivan Rival

Abstract. Properties of several sorts of lattices of convex subsets of ℝ^{n} are examined. The lattice of convex sets containing the origin turns out, for n > 1, to satisfy a set of identities strictly between those of the lattice of all convex subsets of ℝ^{n} and the lattice of all convex subsets of ℝ^{n−1}. The lattices of arbitrary, of open bounded, and of compact convex sets in ℝ^{n} all satisfy the same identities, but the last of these is join-semidistributive, while for n > 1 the first two are not. The lattice of relatively convex subsets of a fixed set S ⊆ ℝ^{n} satisfies some, but in general not all of the identities of the lattice of “genuine” convex subsets of ℝ^{n}.

1. Notation, conventions, remarks

For S a subset of ℝ^{n}, the convex hull of S will be denoted

c.h.(S) = {Σ_{i=1}^{m} λ_{i} p_{i} | m ≥ 1, p_{i} ∈ S, λ_{i} ∈ [0, 1], Σ λ_{i} = 1}.  (1)

When S is written as a list of elements “{ ... },” we shall in general simplify “c.h.({ ... })” to “c.h.( ...)”.

Conv(ℝ^{n}) will denote the lattice of all convex subsets of ℝ^{n}; its lattice operations are

x ∧ y = x ∩ y, x ∨ y = c.h.(x ∪ y). (2)

In any lattice, if a finite family of elements y_{i} has been specified, then an expression such as ⋀_{i} y_{i} will denote the meet over the full range of the index i, and similarly for joins. Likewise, if we write something like ⋀_{j≠i} y_{j} where i has been quantified outside this expression, then the meet will be over all values of j in the indexing family other than i. If L is any lattice and x an element of L, or, more generally, of an overlattice of L, we define the sublattice

L_{≥x} = {y ∈ L | y ≥ x}.  (3)

In particular, Conv(ℝ^{n})_{≥{0}} is the lattice of those convex subsets of ℝ^{n} that contain the origin.

If A is a class of lattices, V(A) will denote the variety of lattices generated by A, that is, the class of lattices satisfying all identities (in the binary operations

1991 Mathematics Subject Classification: Primary: 06B20, 52A20, Secondary: 06E10, 54H12. Key words and phrases: Lattices of convex subsets of ℝ^{n}, n-distributive lattice, meet- and

join-semidistributive lattice, relatively convex set, topologies on power sets of topological spaces. A preprint of this paper is available at arXiv:math.MG/0409288.

∧ and ∨) that hold in all lattices in A. Again, for A given as a list, we will abbreviate V({ ... }) to V( ...). (Most often A will be a singleton {L}, so that we will write V(L).)

Given sets x and y, we will write x − y for their set-theoretic difference, {p | p ∈ x, p ∉ y}.

Though I am not an expert either in lattice theory or in convex sets, I know more about the former subject than the latter; hence, I may more often state explicitly facts known to every worker in convex sets than those known to every lattice-theorist. I hope this note will nevertheless be of interest to people in both fields. I have no present plans of carrying these investigations further; others are welcome to do so.

Since obtaining the main results of this paper, I have learned that many of them were already in the literature, and have added references; thus, this is now a hybrid research/survey paper. I am grateful to Kira Adaricheva, J. B. Nation, Marina Semenova, Fred Wehrung, and the referee for corrections, information on the literature, and many other helpful comments.

Whereas this note looks at conditions satisfied universally in various lattices of convex sets, the papers [1], [19], and others cited there study sufficient conditions for lattices to be embeddable in such lattices, in other words, existential properties of such lattices. (This note also includes one result of that type, in §13.)

2. n-Distributivity

The varieties of lattices we will be examining in the first few sections are those listed in

Lemma 1. Each lattice in the sequence

Conv(ℝ^{0})_{≥{0}}, Conv(ℝ^{0}), Conv(ℝ^{1})_{≥{0}}, …, Conv(ℝ^{n})_{≥{0}}, Conv(ℝ^{n}), …

is embeddable in the next. Hence

V(Conv(ℝ^{0})_{≥{0}}) ⊆ V(Conv(ℝ^{0})) ⊆ V(Conv(ℝ^{1})_{≥{0}}) ⊆ V(Conv(ℝ^{1})) (4)
⊆ ⋯ ⊆ V(Conv(ℝ^{n})_{≥{0}}) ⊆ V(Conv(ℝ^{n})) ⊆ ….

Proof. On the one hand, Conv(ℝ^{n})_{≥{0}} is a sublattice of Conv(ℝ^{n}); on the other, one can embed Conv(ℝ^{n}) in Conv(ℝ^{n+1})_{≥{0}} by sending each x ∈ Conv(ℝ^{n}) to the cone c.h.({0} ∪ {(p_{1}, …, p_{n}, 1) | (p_{1}, …, p_{n}) ∈ x}). The inclusions (4) follow from these embeddings. □

Clearly Conv(ℝ^{0})_{≥{0}} is a trivial lattice with unique element {0}; on the other hand, Conv(ℝ^{0}) is a two-element lattice, so V(Conv(ℝ^{0})) is the variety of distributive lattices. Hence the first inclusion of (4) is strict. We shall see below that the next inclusion is an equality, while all subsequent inclusions are again strict.

Let us set up notation for some relations among elements in a lattice.

Definition 2 (after Huhn [11]). For each positive integer n, we shall denote by D_{n}(x, y_{1}, …, y_{n+1}) (for “ n-distributivity”) the lattice-relation in n+2 arguments x, y_{1}, …, y_{n+1},

x ∧ (⋁_{i} y_{i}) = ⋁_{i} (x ∧ ⋁_{j≠i} y_{j}).  (5)

We shall say that a lattice L satisfies “the identity D_{n}” if (5) holds for all x, y_{1}, …, y_{n+1} ∈ L.

Thus, D_{1} is the ordinary distributivity identity.

The use of the subscript n for an identity in n+2 variables which is symmetric in n+1 of these may seem confusing; a useful mnemonic is that D_{n} is the identity that allows one to “reduce meets of x with larger joins to meets of x with n-fold joins”. An additional occasion for confusion will arise when we see that the first of these identities to be satisfied by Conv(ℝ^{n}) is not D_{n}, but D_{n+1}. This will be a consequence of the fact that an n-dimensional simplex has n+1 vertices.

Note that the left-hand side of (5) is ≥ the right-hand side for any family of elements of any lattice, since each of the n+1 terms in the outer join on the right is majorized by the left-hand side. So to verify any instance of (5) it suffices to prove “ ≤”.

The pioneering work on identities satisfied by lattices of convex sets was done by A. P. Huhn [10], [11]. The results in this and the next three sections will extend Huhn’s by approximately doubling both the family of lattices and the family of identities considered, and formalizing some general techniques. Huhn’s results will be recovered along with our new ones.

The key to Huhn’s and our results on D_{n} (and some related identities) is the following standard result in the theory of convex sets. Strictly speaking, it is the first sentence below that is Carathéodory’s theorem, while the second is a wellknown refinement thereof [6, p.431, line 4]. Intuitively, that second sentence says that from an arbitrary point p_{0} ∈ S ⊆ ℝ^{n}, we can “see” any other point of c.h.(S) against a background of (or embedded in) some (n−1)-simplex with vertices in S.

Carathéodory’s Theorem. If S is a subset of ℝ^{n}, then each element q ∈ c.h.(S) belongs to c.h.(p_{0}, …, p_{n}) for some n+1 points p_{0}, …, p_{n} ∈ S. Moreover, p_{0} can be taken to be any pre-specified element of S.

In each paragraph of the next lemma, it is the first assertion that is due to Huhn.

Lemma 3 (cf. Huhn [11]). For every natural number n and (n+3)-tuple of convex sets x, y_{1}, …, y_{n+2} ∈ Conv(ℝ^{n}), the relation D_{n+1}(x, y_{1}, …, y_{n+2}) holds. Moreover, if n > 0 and (at least) some two of the sets y_{1}, …, y_{n+1} have nonempty intersection, then D_{n}(x, y_{1}, …, y_{n+1}) holds.

Hence Conv(ℝ^{n}) satisfies the identity D_{n+1}, and for n > 0, Conv(ℝ^{n})_{≥{0}} satisfies the identity D_{n}.

Proof. We shall prove the assertions of the first paragraph, which clearly imply those of the second.

To get the first assertion, let x, y_{1}, …, y_{n+2} ∈ Conv(ℝ^{n}), and let p be a point of the convex set described by the left-hand side of (5). Then p belongs to both x and ⋁_{i} y_{i} = c.h.(⋃_{i} y_{i}). Carathéodory’s Theorem now says that p belongs to the convex hull of n+1 points of ⋃_{i} y_{i}, hence to the join of at most n+1 of the y_{i}, so it belongs to one of the terms on the right-hand side of (5), hence to their join, as required.

To prove the second assertion, let x, y_{1}, …, y_{n+1} ∈ Conv(ℝ^{n}), and suppose that two of y_{1}, …, y_{n+1} have a point p_{0} in common. Given p in the left-hand side of the desired instance of (5), we see from the last sentence of our statement of Carathéodory’s Theorem that p will belong to the convex hull of p_{0} and some n other points of ⋃_{i} y_{i}. The latter points will lie in the union of some n of the y ’s, and since those n of our n+1 sets leave out only one, they do not leave out both of the sets known to contain p_{0}. So that family of n y ’s contains the n+1 points whose convex hull is known to contain p, and the conclusion follows as before. □

Note that for all p ∈ ℝ^{n}, we have Conv(ℝ^{n})_{≥{p}} ≅ Conv(ℝ^{n})_{≥{0}}; moreover, if p ∈ S ⊆ ℝ^{n}, then Conv(ℝ^{n})_{≥S} ⊆ Conv(ℝ^{n})_{≥{p}}. Thus, any identities proved for the lattice Conv(ℝ^{n})_{≥{0}} will hold in Conv(ℝ^{n})_{≥S} for every nonempty set S.

One may ask whether, in the first assertion of the above lemma, we have failed to use the full strength of Carathéodory’s Theorem. That theorem says that the convex hull of any family of N > n+1 points is the union of the hulls of its (n+1)- element subfamilies, hence for each such N, we may deduce an identity like D_{n+1}, but with the join of N rather than n+2 convex sets y_{i} on the left, and expressions involving all the (n+1)-fold subjoins thereof on the right. Our D_{n+1} is the case N = n+2.

But in fact, the identities so obtained are all equivalent to D_{n+1}. To derive them from it, note first that if in D_{n+1} we substitute for y_{n+2} the expression y_{n+2}∨y_{n+3}, then the left-hand side becomes x∧(⋁_{i=1}^{n+3} y_{i}), while on the right, some of the joins involve n+1 of the y ’s and others involve n+2. If we again apply D_{n+1} to the latter joins, we get precisely the N = n+3 case of the class of identities discussed above. The identities with still larger N are gotten by repeating this argument. Conversely, one can get D_{n+1} from any of these identities by substituting for the y_{i} with i > n+2 repetitions of y_{n+2}, and discarding from the outer join on the right-hand side joinands majorized by others.

Let us also note that D_{n+1} ⟹ D_{n+2}. Indeed, the identity with n+3 y ’s that we just showed equivalent to D_{n+1} has the same left-hand side as D_{n+2}, while the right-hand side of D_{n+2} can be seen to lie, in an arbitrary lattice, between the two sides of that identity. In particular, the identities obtained in Lemma 3 are successively weaker for larger n, as is reasonable in view of (4).

The argument proving Lemma 3 (for simplicity let us limit ourselves to the first assertion thereof) can be formulated in a more general context, and the above observations on identities allow us to obtain a converse in that context, which we record below, though we shall not use it. Recall that a closure operator on a set X is called “finitary” (or “algebraic”) if the closure of every subset S ⊆ X is the union of the closures of the finite subsets of S.

Lemma 4 (cf. [16]). Let cl be a finitary closure operator on a set X, such that every singleton subset of X is closed, or more generally, such that the closure of every singleton is finitely join-irreducible; and let n be a positive integer. Then the lattice of closed subsets of X under cl satisfies D_{n} if and only if cl has the “ n-Carathéodory property” that the closure of every set is the union of the closures of its ≤n-element subsets.

Proof. “If” is shown exactly as in the proof of Lemma 3.

Conversely, suppose the lattice of cl-closed subsets of X satisfies D_{n}, and let p ∈ cl(S) for some S ⊆ X. By the assumption that cl is finitary, p ∈ cl(q_{1}, …, q_{N}) for some q_{1}, …, q_{N} ∈ S. If N ≤ n, we are done; if not, let us rewrite the condition p ∈ cl(q_{1}, …, q_{N}) as a lattice relation, cl(p) = cl(p) ∧ (⋁_{i} cl(q_{i})). By the preceding discussion, D_{n} implies the identity in N variables which, when applied to the right-hand side of the above relation, turns the relation into

cl(p) = ⋁_{|I|=n} (cl(p) ∧ (⋁_{i∈I} cl(q_{i}))),

where the subscript to the outer join means that I ranges over all n-element subsets of {1, …, N}. Now by assumption cl(p) is finitely join-irreducible, hence it equals one of the joinands on the right, showing p to be in the closure of some set of n q ’s, as required. □

3. Tools for studying related identities

We also appear to have used less than the full force of the middle sentence of Lemma 3 (the stronger relation holding when at least two of the y ’s have nonempty intersection) in getting the second assertion of the last sentence of that lemma (the stronger identity for Conv(ℝ^{n})_{≥{0}}), since the latter assertion concerns the case where not just two, but all the y_{i} (and also x) have a point in common. There is no evident way to take advantage of the weaker hypothesis of said middle sentence when working in the lattice Conv(ℝ^{n})_{≥{0}}; but might we be able to use it to prove some new identity for Conv(ℝ^{n}); for example, some relation that we can show holds, on the one hand, whenever a family of elements x, y_{1}, …, y_{n+1} satisfies D_{n}, and also, for some trivial reason, whenever y_{1} ∧ y_{2} = ∅?

One relation with these properties can be obtained by taking the meet of each side of D_{n} with y_{1} ∧ y_{2}. Unfortunately, this turns out to be the trivial identity: both sides simplify to x ∧ y_{1} ∧ y_{2} in any lattice. However, we can circumvent this by first taking the join of both sides of D_{n} with a new indeterminate z, and only then taking meets with y_{1} ∧ y_{2}.

We shall in fact see that this provides what Lemma 3 failed to: an identity holding in Conv(ℝ^{n}) but not in Conv(ℝ^{n+1})_{≥{0}}. However, the verification of an example showing the failure of this identity in the larger lattice (and of a similar example we will need later) is messy if done entirely “by hand”; so we shall establish in this section some general criteria for certain sorts of inequalities in lattices of convex sets to be strict.

The last assertions of the next two lemmas clearly imply, for Conv(ℝ^{n}) and Conv(ℝ^{n+1})_{≥{0}} respectively, that if we are given expressions u, v and w in some lattice indeterminates such that the inequality u ≥ v is known to hold identically in our lattice, then we can write down another inequality holding identically among expressions in a slightly larger set of indeterminates, for which equality will hold whenever either u equals v, or w equals the least lattice-element (i.e., ∅, respectively {0}), but which will fail in all other cases. This is what is logically called for by the program sketched above. However the earlier parts of these lemmas give some simpler inequalities for which the same is true if the hypotheses hold “in a sufficiently strong way”, and it will turn out that in the applications where we need to show failure of an identity, we will be able to use these simpler formulas.

In the statements of these lemmas, for p, q ∈ ℝ^{n}, “the ray drawn from q through p” will mean {λp + (1 − λ)q | 0 ≤ λ < ∞}, even in the degenerate case p = q, where this set is the singleton {p}.

Lemma 5. Let n be any natural number, and let u ≥ v and w be three elements of Conv(ℝ^{n}). Then

(i) The following conditions are equivalent:

(a) There exists z ∈ Conv(ℝ^{n}) such that (u ∨ z) ∧ w > (v ∨ z) ∧ w.

(b) There exist a point p ∈ u and a point q ∈ w such that the ray drawn from q through p contains no point of v. (ii) The following conditions are equivalent:

(a) There exist z, z′ ∈ Conv(ℝ^{n}) such that ((u ∧ z′) ∨ z) ∧ w > ((v ∧ z′) ∨ z) ∧ w.

(b) u > v, and w is nonempty.

Proof. In view of the hypothesis u ≥ v, the relation “ ≥” always holds in the inequalities of (i)(a) and (ii)(a), so in each case, strict inequality is equivalent to the existence of a point q belonging to the left-hand side but not to the right-hand side.

Suppose, first, that (i)(a) holds for some z; thus we get a point q belonging to u∨z and to w but not to v ∨z. Note that the latter condition implies that q does not lie in z. If q lies in u, then (i)(b) is satisfied with p = q, so assume q ∉ u. Hence, being in the convex hull of u and z but in neither set, q must lie on the line-segment c.h.(p, r) for some p ∈ u, r ∈ z :

[figure: p; ∈ u; q; ∈ (u ∨ z) ∧ w; ∉ v ∨ z; r; ∈ z]

Now if a point s ∈ v lay on the ray drawn from q through p, we would have q ∈ c.h.(s, r) ⊆ v ∨z, contradicting our assumption that q ∉ v ∨z. This proves (b).

Conversely, if we are given p and q as in (i)(b), take z = {2q − p} :

[figure: p; ∈ u; q; ∈ w; 2q − p; ∈ z]

Thus q ∈ (u ∨ z) ∧ w, but we claim that q ∉ v ∨ z. Indeed, if q belonged to this set, then v would have to meet the ray from the unique point 2q − p of z through q on the other side of q; i.e., it would meet the ray drawn from q through p, contradicting our choice of p and q as in (b). Hence q belongs to the left-hand but not the right-hand side of the inequality of (a), as required.

Note that (i)(b), and hence (i)(a), holds whenever u and w are nonempty and v is empty.

Now in the situation of (ii)(b), if we take any p ∈ u − v and let z′ = {p}, then u ∧ z′ and v ∧ z′ are respectively nonempty and empty, so applying the preceding observation with these two sets in the roles of u and v, we get (ii)(a). The reverse implication is trivial. □

The result we shall prove for Conv(ℝ^{n})_{≥{0}} is similar. Indeed, in the lemma below, part (i) is exactly as in the preceding lemma; but (ii) becomes two statements, (ii) and (iii), the former having the same “(a)” as in (ii) above but a stronger “(b)”, the latter a weaker “(a)” but essentially the same “(b)” as above. Note also the restriction on n (only needed for (iii)).

Lemma 6. Let n > 1, and let u ≥ v and w be three elements of Conv(ℝ^{n})_{≥{0}}. Then

(i) The following conditions are equivalent:

(a) There exists z ∈ Conv(ℝ^{n})_{≥{0}} such that (u ∨ z) ∧ w > (v ∨ z) ∧ w.

(b) There exist a point p ∈ u and a point q ∈ w such that the ray drawn from q through p contains no point of v. (ii) The following conditions are equivalent:

(a) There exist z, z′ ∈ Conv(ℝ^{n})_{≥{0}} such that

((u ∧ z′) ∨ z) ∧ w > ((v ∧ z′) ∨ z) ∧ w.

(b) There exist a point p ∈ u and a point q ∈ w such that the ray drawn from q through p contains no point of v ∧ c.h.(0, p). (iii) The following conditions are equivalent:

(a) There exist z, z′, z′′, z′′′ ∈ Conv(ℝ^{n})_{≥{0}} such that

((((u ∧ z′′′) ∨ z′′) ∧ z′) ∨ z) ∧ w > ((((v ∧ z′′′) ∨ z′′) ∧ z′) ∨ z) ∧ w.

(b) u > v, and w ≠ {0}.

Proof. (i)(a) ⟹ (i)(b) holds by the preceding lemma. In proving the reverse implication, we cannot set z = {2q − p} as we did there, so let z = c.h.(0, 2q − p). As before, we have q ∈ (u ∨ z) ∧ w and need to show q ∉ v ∨ z. If the contrary were true, then q would be a convex linear combination of 0, 2q − p, and a point r ∈ v. This can be rewritten as a convex linear combination of 2q−p with a convex linear combination of 0 and r; but the latter combination would also be a point of v, and, as in the previous proof, would lie on the ray drawn from q through p, contradicting our choice of p and q as in (b).

Turning to (ii), if (ii)(a) holds then we can apply (i)(a) ⟹ (i)(b) with u ∧ z′ and v ∧ z′ in place of u and v, and the resulting p and q will satisfy (ii)(b) (since v ∧ c.h.(0, p) ⊆ v ∧ z′). Inversely, if (ii)(b) holds, take z′ = c.h.(0, p), and apply (i)(b) ⟹ (i)(a) with u ∧ z′ and v ∧ z′ in place of u and v.

In statement (iii), it is clear that (a) implies (b). To prove the converse, let us assume (iii)(b), and consider two cases, according to whether the stronger statement (ii)(b) holds. If it does, we get the inequality of (ii)(a), from which we can immediately get that of (iii)(a) by choosing z′′′ and z′′ to “have no effect” (e.g., by taking them to be u and v respectively). On the other hand, if (ii)(b) fails while (iii)(b) holds, it is easy to see that all elements of w and u − v must lie on a common line x through 0. In that case, we want to use z′′′ and z′′ to “perturb” u and v, so that the modified u has points off the line x, while being careful to preserve the property that u is strictly larger than v. To do this, we begin by taking any point p ∈ u − v, letting z′′′ = c.h.(0, p), and noting that the intersections of u and v with this segment are still distinct. Now taking any point r not on the line x (it is for this that we need n > 1), and letting z′′ = c.h.(0, r), we see that (u ∧ z′′′) ∨ z′′ and (v ∧ z′′′) ∨ z′′ remain distinct, and that their difference now has points off x. Hence (ii)(b) holds with these sets in the roles of u and v, and the implication (ii)(b) ⟹ (ii)(a) gives the z and z′ needed for (iii)(a). □

Digression: One can get criteria similar to those of the preceding lemmas for other conditions. At the trivial end, given u ∈ Conv(ℝ^{n}), a condition for u to be nonempty is that there exist a z such that u ∨ z > z, and likewise the condition for at least one of two elements u and v to be nonempty is that their join have this property. The exercise below offers, for the diversion of the interested reader, some less trivial cases.

Exercise 7. (i) Find an inequality in elements u, v and one or more additional lattice variables z, …, which holds identically in lattices, and which, for any u, v ∈ Conv(ℝ^{n}), is strict for some values of the additional variables if and only if both u and v are nonempty. (ii) Find an inequality in elements u, v, w, x and additional variables which holds in any lattice when u ≥ v, w ≥ x, and which, for any such u, v, w, x ∈ Conv(ℝ^{n}), is strict for some values of the additional variables if and only if u > v and w > x. (iii) Same as (ii), but with “ u > v and w > x” replaced by “ u > v or w > x”. (iv)-(vi) Like (i)-(iii), but for Conv(ℝ^{n})_{≥{0}}, and with “nonzero” in place of “nonempty” in (i). (vii) In Lemma 6, conditions (i)(a), (ii)(a) and (iii)(a) involved 1, 2 and 4 z’s respectively; so the condition that there exist three elements z, z′, z′′ ∈ Conv(ℝ^{n})_{≥{0}} such that

(((u ∨ z′′) ∧ z′) ∨ z) ∧ w > (((v ∨ z′′) ∧ z′) ∨ z) ∧ w

was skipped. Show by example that this condition is not equivalent to (iii)(b). (viii) Suppose u ≥ v and w are elements of Conv(ℝ^{n}), and consider the conditions dual to (i)(a)-(iii)(a) of Lemma 6, and to the “skipped” condition:

(a) There exists z ∈ Conv(ℝ^{n}) such that (u ∧ z) ∨ w > (v ∧ z) ∨ w.

(a′) There exist z, z′ ∈ Conv(ℝ^{n}) such that ((u∨z′)∧z)∨w > ((v ∨z′)∧z)∨w. (a′′) There exist z, z′, z′′ ∈ Conv(ℝ^{n}) such that

(((u ∧ z′′) ∨ z′) ∧ z) ∨ w > (((v ∧ z′′) ∨ z′) ∧ z) ∨ w.

(a′′′) There exist z, z′, z′′, z′′′ ∈ Conv(ℝ^{n}) such that

((((u ∨ z′′′) ∧ z′′) ∨ z′) ∧ z) ∨ w > ((((v ∨ z′′′) ∧ z′′) ∨ z′) ∧ z) ∨ w.

Which of these, if any, are equivalent, for all such u, v and w, to the condition (b) u > v, and w ≠ ℝ^{n}?

Now, back to business.

4. Identities distinguishing our chain of lattices

Given n > 0, let us write ((D_{n}) ∨ z) ∧ y_{1} ∧ y_{2} for the equation in n+3 variables x, y_{1}, …, y_{n+1}, z obtained by applying the operator ((−) ∨ z) ∧ y_{1} ∧ y_{2} to both sides of the relation D_{n}(x, y_{1}, …, y_{n+1}). We can now prove

Theorem 8. For each positive integer n, Conv(ℝ^{n}) satisfies ((D_{n})∨z)∧y_{1}∧y_{2} but not D_{n}, while Conv(ℝ^{n})_{≥{0}} satisfies D_{n} but, if n > 1, not ((D_{n−1})∨z)∧y_{1}∧y_{2}.

Hence every inclusion in (4) is strict except the second. Equality holds at that step.

Proof. To see that equality holds at the second inclusion of (4), note that both Conv(ℝ^{0}) and Conv(ℝ^{1})_{≥{0}} are nontrivial lattices, which by Lemma 3 satisfy D_{1}, the distributive identity, and that the variety of distributive lattices is known to have no proper nontrivial subvarieties.

The positive assertions of the theorem’s first paragraph follow from Lemma 3, combined, in the case of the first of these results, with the observations of the first two paragraphs of §3. (Those observations are equivalent to the contrapositive of the easy implications (i)(a) ⟹ (ii)(a) ⟹ (ii)(b) of Lemma 5.) It remains to give examples showing the negative assertions.

To see that Conv(ℝ^{n}) does not satisfy D_{n}, let q_{1}, …, q_{n+1} be the vertices of an n-simplex in ℝ^{n} and p an interior point of this simplex, and take for the y_{i} and x the singletons {q_{i}} and {p} respectively. Then we see that the left-hand side of D_{n}(x, y_{1}, …, y_{n+1}) gives x, while all the joinands on the right are empty, hence so is the right-hand side itself.

To show that Conv(ℝ^{n})_{≥{0}} does not satisfy ((D_{n−1})∨z)∧y_{1} ∧y_{2} when n > 1, let P ⊆ ℝ^{n} be a hyperplane not passing through 0. The idea will be to mimic the preceding example within P, then replace the resulting singleton sets with the line-segments connecting them with 0, slightly enlarge y_{1}, so that it has nonzero intersection with y_{2}, and finally apply part (i) of Lemma 6.

So let q_{1}, …, q_{n} be the vertices of an (n−1)-simplex in P, and p a point in the relative interior of that simplex, and let x and the y_{i} be the line segments c.h.(0, p) and c.h.(0, q_{i}) respectively, except for y_{1}, which we take to be c.h.(0, q_{1}, q_{2}/2). Note that all of these convex sets lie in the closed half-space H bounded by P and containing 0; hence the intersection with P of any lattice expression in these convex sets can be computed as the corresponding lattice expression in their intersections with P. We see that these intersections are a configuration of the form given in the preceding example, except that the dimension is lower by 1. (Note that the point “ q_{2}/2” in our definition of y_{1}, not lying in P, does not affect the intersection y_{1} ∩ P.) Hence when we evaluate the two sides of D_{n−1} at these elements, the left-hand side intersects P in the point p, and is thus the whole line-segment x, while the right-hand side does not meet P, hence is a proper subsegment of x.

We can now deduce the failure of ((D_{n−1}) ∨ z) ∧ y_{1} ∧ y_{2} from the implication (i)(b) ⟹ (i)(a) of Lemma 6, using for u and v respectively the left and right sides of D_{n−1}, and for w the set y_{1}∧y_{2}. To see that (i)(b) holds for these sets, note that w is the line-segment c.h.(0, q_{2}/2). Since this lies in a different line through the origin from p, the ray drawn from any nonzero point q of that segment through p does not meet the line-segment x in any point other than p, so in particular, it does not contain any point of v, the right-hand side of D_{n−1}, which we saw was a proper subset of x. □

We remark that by alternately inserting joins and meets with more and more variables z^{(m)} into “((D_{n}) …)∧(y_{1} ∧y_{2})” one can get identities that, formally, are successively stronger (though still all implied by D_{n}), so that the statements that a lattice does not satisfy these identities become successively weaker. Thus, as a stronger version of the above theorem, we could have stated that Conv(ℝ^{n}) satisfies such identities with arbitrarily long strings of inserted terms, while Conv(ℝ^{n})_{≥{0}} fails to satisfy the particular one given above. But for simplicity, I used just one identity to distinguish the properties of these lattices.

The argument at the beginning of the preceding section showing that Conv(ℝ^{n}) satisfies ((D_{n})∨z)∧y_{1} ∧y_{2} also clearly shows that it satisfies the formally stronger identity

((D_{n}) ∨ z) ∧ (⋁_{i,j; i ≠ j} y_{i} ∧ y_{j}).  (6)

With a little additional work one can get the still stronger identity:

((D_{n}) ∨ z) ∧ ⋀_{i} (⋁_{j≠i} y_{j}).  (7)

(Idea: If ⋀_{i} (⋁_{j≠i} y_{j}) is nonempty, use a point thereof as the p_{0} in the second sentence of Carathéodory’s Theorem.)

We remark that the fact that Conv(ℝ^{n}) satisfies the identity D_{n+1} says that for any n+2 convex sets y_{i}, the union of the (n+1)-fold joins ⋁_{j≠i} y_{j} (i = 1, …, n+2) is itself convex, while (7) says essentially that the same holds for the union of the n-fold joins of n+1 convex sets, if those joins have at least one point in common.

5. Dual n-distributivity

Huhn showed not only that Conv(ℝ^{n}) satisfies D_{n+1}, but also that it satisfies the dual of that identity. Let us write the dual of the relation D_{n}(x, y_{1}, …, y_{n+1}) as

D_{n}^{op}(x, y_{1}, …, y_{n+1}) : x ∨ (⋀_{i} y_{i}) = ⋀_{i} (x ∨ ⋀_{j≠i} y_{j}).  (8)

Just as in D_{n} the direction ≥ is automatic, so ≤ is automatic in D_{n}^{op}.

Below, we will strengthen Huhn’s result that D_{n+1}^{op} holds identically in Conv(ℝ^{n}) by showing that it holds in the larger lattice Conv(ℝ^{n+1})_{≥{0}}, and will again use the method of §3 to manufacture a related identity which holds in Conv(ℝ^{n}) but not in Conv(ℝ^{n+1})_{≥{0}}. Like Huhn, we start with

Helly’s Theorem ([6], p.391). Let n ≥ 0. If a finite family of convex subsets of ℝ^{n} has the property that every n+1 of them have nonempty intersection, then the whole family has nonempty intersection.

We will also use the following observation.

Lemma 9. Let V be any real vector space. Given a convex set x and a point p in V, there exists a convex set w in V such that for every nonempty convex set y, one has p ∈ x ∨ y if and only if y has nonempty intersection with w.

Proof. It is straightforward to check that a set w with the required property (in fact, the unique such set) is the cone consisting of the union of all rays from p which meet the central reflection of x through p. (If p ∈ x then w = V; in visualizing the contrary case, it is convenient to assume without loss of generality that p = 0.) □

We can now show that certain sorts of families of convex sets satisfy the relation D_{n}^{op}, from which we will deduce our identities.

Lemma 10. Let n be a natural number, and x, y_{1}, …, y_{n+1} be n+2 elements of Conv(ℝ^{n}) such that y_{1} ∧ ⋯ ∧ y_{n+1} ≠ ∅. Then D_{n}^{op}(x, y_{1}, …, y_{n+1}) holds.

Proof. Given

p ∈ ⋀_{i} (x ∨ ⋀_{j≠i} y_{j})  (9)

we need to show that

p ∈ x ∨ (⋀_{i} y_{i}).  (10)

Let w be the set determined by x and p as in Lemma 9. Since by (9), p lies in each of the sets x∨⋀_{j≠i} y_{j}, our choice of w shows that for each i, (⋀_{j≠i} y_{j})∧w is nonempty. These conditions together with the nonemptiness of y_{1}∧⋯∧y_{n+1} allow us to apply Helly’s Theorem to the n+2 convex sets y_{1}, …, y_{n+1}, w, and conclude that (⋀_{i} y_{i}) ∧ w is nonempty, which by choice of w is equivalent to (10). □

Now for each positive integer n, let (((D_{n}^{op}) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}) denote the relation in variables x, y_{1}, …, y_{n+1}, z, z′ obtained by applying the operation (((−) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}) to both sides of (8). Then we have

Theorem 11 (cf. Huhn [10]). For each positive integer n, Conv(ℝ^{n}) satisfies the identity (((D_{n}^{op}) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}) but not D_{n}^{op}, while Conv(ℝ^{n})_{≥{0}} satisfies the identity D_{n}^{op} but, if n > 1, not (((D_{n−1}^{op}) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}).

Proof. The positive assertions follow from Lemma 10. (Again, the reasoning that obtains the first identity from that lemma can be considered, formally, an application of the contrapositive of an easy direction in of one of our lemmas, in this case the implication (ii)(a) ⟹ (iii)(b) of Lemma 6.)

To get a counterexample to D_{n}^{op} in Conv(ℝ^{n}), take for x an n-simplex in ℝ^{n}, let p be a point outside x, let x′ be the central reflection of x through p, and let y_{1}, …, y_{n+1} be the (n−1)-faces of x′. Then ⋀_{i} y_{i} is empty, so the left-hand side of (8) is just x, while each of the intersections ⋀_{j≠i} y_{j} is a nonempty subset of x′, so the right-hand side of (8) contains p.

As in the proof of Theorem 8, we begin the counterexample to our more elaborate identity in Conv(ℝ^{n})_{≥{0}} by taking a copy of our preceding example, for the next lower dimension, in a hyperplane P ⊆ ℝ^{n} not containing 0. Let us write p_{0} and x_{0}, x′_{0}, y_{10}, …, y_{n0} for the point and family of convex subsets of P so obtained. Let us also write q_{0} for the vertex of x′_{0} opposite to the face y_{10}.

To beef these sets up to the desired members of Conv(ℝ^{n})_{≥{0}}, we now take

x = c.h.(x_{0} ∪ {0}), y_{i} = c.h.(y_{i0} ∪ {0}) for i ≠ 1, y_{1} = c.h.(y_{10} ∪ {0, q_{0}/2}).

For the same reason as in the proof of Theorem 8, the operation of intersecting with P commutes with lattice operations on these convex sets; hence when D_{n−1}^{op} is evaluated at the above arguments, the left-hand side meets the plane P only in the set x_{0}, while the right-hand side will also contain the point p_{0}. Now observe that ⋀_{i} y_{i} will be the line-segment c.h.(0, q_{0}/2). Letting p = p_{0}, q = q_{0}/2, we see that these lie on different lines through 0, and deduce from the implication (ii)(b) ⟹ (ii)(a) of Lemma 6 that z and z′ can be chosen so that the required inequality holds. □

6. Encore!

Carathéodory’s and Helly’s Theorems are two members of a well-known triad of results on convex sets in ℝ^{n}. The third result is

Radon’s Theorem ([6], p.391). Given a natural number n, and n+2 points p_{1}, …, p_{n+2} in ℝ^{n}, there exists a partition of {1, …, n+2} into subsets I_{1} and I_{2} such that c.h.({p_{i} | i ∈ I_{1}}) ∩ c.h.({p_{i} | i ∈ I_{2}}) ≠ ∅.

Can we turn this, too, into an identity for lattices of convex sets?

Yes. First let’s get rid of reference to points: Clearly an equivalent statement is “Given nonempty convex sets y_{1}, …, y_{n+2} in ℝ^{n}, there exists a partition of {1, …, n+2} into subsets I_{1} and I_{2} such that (⋁_{i∈I_{1}} y_{i}) ∧ (⋁_{i∈I_{2}} y_{i}) ≠ ∅.” Next, the conclusion that such a partition exists can be condensed into the single inequality ⋁_{I_{1},I_{2}} ((⋁_{i∈I_{1}} y_{i}) ∧ (⋁_{i∈I_{2}} y_{i})) ≠ ∅, where the outer join is over the 2^{n+2} − 2 partitions of {1, …, n+2} into two nonempty subsets. In the proof of the theorem below, Lemma 9 will be used to turn the above implication between nonemptiness statements into a lattice relation.

In that theorem, I call an inequality a ≤ b that always holds an “identity”, since it can be rewritten as an equation a = a ∧ b.

Theorem 12. For every natural number n, the lattice identity in n+3 variables x, y_{1}, …, y_{n+2}

⋀_{i} (x ∨ y_{i}) ≤ x ∨ ⋁_{I_{1},I_{2}}((⋁_{i∈I_{1}} y_{i}) ∧ (⋁_{i∈I_{2}} y_{i})),  (11)

where “ ⋁_{I_{1},I_{2}}” denotes the join over all partitions of {1, …, n+2} into two nonempty subsets I_{1} and I_{2}, holds in Conv(ℝ^{n}), and indeed in Conv(ℝ^{n+1})_{≥{0}}, but not in Conv(ℝ^{n+1}).

Proof. We shall first prove (11) in the simpler case of Conv(ℝ^{n}), then show how to adapt the proof to Conv(ℝ^{n+1})_{≥{0}}, and finally, give the counterexample in Conv(ℝ^{n+1}).

Given x, y_{1}, …, y_{n+2} ∈ Conv(ℝ^{n}) and a point p belonging to the left-hand side of (11), we must show that p also belongs to the right-hand side. Let us choose w as in Lemma 9 for the given x and p. The assumption that p belongs to the left-hand side of (11) says that it belongs to each of the meetands of that expression, which by choice of w means that w ∧ y_{i} is nonempty for all i. Hence Radon’s Theorem applied to those n+2 sets says that for some I_{1}, I_{2} partitioning {1, …, n+2}, we have ∅ ≠ (⋁_{I_{1}} w ∧ y_{i}) ∧ (⋁_{I_{2}} w ∧ y_{i}). The latter set is contained in w ∧ (⋁_{I_{1}} y_{i}) ∧ (⋁_{I_{2}} y_{i}), and the statement that this is nonempty now translates back to say that x ∨ ((⋁_{I_{1}} y_{i}) ∧ (⋁_{I_{2}} y_{i})) contains p, whence p belongs to the right-hand side of (11), as required.

If we are given x, y_{1}, …, y_{n+2} ∈ Conv(ℝ^{n+1})_{≥{0}}, and a point p on the lefthand side of (11), we begin in the same way, translating the hypothesis and desired conclusion to the same statements about the sets w ∧ y_{i} (though the convex set w will not in general belong to Conv(ℝ^{n+1})_{≥{0}}, and so neither will these intersections). This time we apply Radon’s Theorem in n+1 dimensions to the sets w ∧ y_{i} together with {0}. In the partition given by that theorem, let us assume without loss of generality that {0} goes into the second join; thus we get a relation ∅ ≠ (⋁_{I_{1}} w ∧ y_{i}) ∧ ({0} ∨ ⋁_{I_{2}} w ∧ y_{i}). Here the first join is contained in w, hence so is the whole set, so that set is contained in the meet of w with the larger set (⋁_{I_{1}} y_{i}) ∧ ({0} ∨ ⋁_{I_{2}} y_{i}). Since all the y ’s belong to Conv(ℝ^{n+1})_{≥{0}}, the joinand {0} is now redundant, so we have again shown that w ∧ (⋁_{I_{1}} y_{i}) ∧ (⋁_{I_{2}} y_{i}) is nonempty for some partition {1, …, n+2} = I_{1} ∨ I_{2}, which, as before, yields the desired conclusion by choice of w.

To show that (11) does not hold in Conv(ℝ^{n+1}), we begin essentially as in the first counterexample in the proof of Theorem 11, letting x be an (n+1)-simplex in ℝ^{n+1}, p a point outside that simplex, and x′ the central reflection of x through p. This time, however, we let y_{1}, …, y_{n+2} be singletons, whose unique elements are the vertices of x′. As in the earlier example we find that one side of the identity in question (in this case the left-hand side of (11)) contains p, while the other is simply x, since all joinands in the “big join” on that side are empty; so the right-hand side does not majorize the left-hand side. □

I have not tried to fill in this picture, as I did with the identities D_{n} and D_{n}^{op}, by looking for related identities that would separate all distinct terms of (4); but I expect that these exist.

Another observation which I have not followed up on, because it occurred to me late in the preparation of this paper, is that if one defines Conv(ℝ^{n}, cone) ⊆ Conv(ℝ^{n})_{≥0} to be the sublattice consisting of those convex sets which are unions of rays through 0, then Conv(ℝ^{n−1}) embeds naturally in Conv(ℝ^{n}, cone), yielding a refinement of the chain of varieties of Lemma 1. It would be interesting to know whether for n > 1 the varieties so interpolated are distinct from those that precede and follow them.

7. The sublattice of compact convex sets

From Carathéodory’s Theorem, we see that

For any compact subset S ⊆ ℝ^{n}, c.h.(S) is also compact.  (12)

Hence the join in Conv(ℝ^{n}) of two compact subsets is compact, hence the set of compact convex subsets of ℝ^{n} (often called “convex bodies” in the literature, e.g., [9], [1, §3.1], [19, §12]) is a sublattice Conv(ℝ^{n}, cpct) ⊆ Conv(ℝ^{n}). An obvious question is how the identities of this sublattice compare with those of Conv(ℝ^{n}); i.e., whether it satisfies any identities that the larger lattice does not. Huhn [11, proof of Lemma 3.1] answered this question in the negative, by showing that the still smaller lattice of polytopes (convex hulls of finite sets) does not. Let me give a slightly different proof of the same result. Huhn used the fact that an intersection of polytopes is a polytope, but the next result is applicable to a finitary closure operator that need not have the property that an intersection of closures of finite sets is again one. (The meaning of “finitary” was recalled in the paragraph preceding Lemma 4; the notation V( ...) used below was defined in §1.)

Proposition 13. Let cl be a finitary closure operator on a set X, and let L be the lattice of subsets of X closed under cl. Then every lattice relation satisfied by all families of elements of L that are closures of finite subsets of X is an identity of L. In particular, if L′ is any sublattice of L which contains all closures of singleton subsets of X, then V(L′) = V(L).

Proof. Let us topologize the power set 2^{X} by taking as a basis of open sets the sets

U(A, B) = {Y ∈ 2^{X} | A ⊆ Y ⊆ B},

where A ranges over the finite subsets of X, and B over arbitrary subsets. This is stronger than the usual power-set topology, which only uses the sets of the above form with B cofinite. Thus, our topology is Hausdorff, though not in general compact; hence its restriction to L is also Hausdorff, with basis of open sets given by the sets U_{L}(A, B) = U(A, B)∩L. (Of course, U_{L}(A, B) is nonempty only when cl(A) ⊆ B.)

We claim that under this topology, the lattice operations of L are continuous, and the closures of finite subsets of X are dense in L. This will imply that any lattice identity holding on that dense subset must hold on all of L, from which the final conclusion will clearly follow.

To see that closures of finite sets are dense, note that every nonempty basic set U_{L}(A, B) contains the element cl(A), which is such a set.

The continuity of the meet operation is also straightforward: If x, y ∈ L are such that x ∧ y, i.e., x ∩ y, lies in U_{L}(A, B), then U_{L}(A, x ∪ B) and U_{L}(A, y ∪ B) are neighborhoods of x and y respectively such that the intersection of any member of the first neighborhood and any member of the second lies in U_{L}(A, B).

Finally, suppose x, y ∈ L are such that x ∨ y, i.e., cl(x ∪ y), lies in U_{L}(A, B). Then the finite set A is contained in cl(x ∪ y), so by finitariness of cl, there is a finite subset of x ∪ y whose closure contains all elements of A; let us write this subset as A_{x} ∪ A_{y} where A_{x} ⊆ x and A_{y} ⊆ y. Then U_{L}(A_{x}, x) and U_{L}(A_{y}, y) will be neighborhoods of x and y respectively such that the join of any member of the first neighborhood and any member of the second is a member of U_{L}(A, B). (In the power-set topology, ∨ is generally discontinuous; this is why we needed a different topology.) □

For X = ℝ^{n} and cl = c.h., the L of the above proposition is Conv(ℝ^{n}). Since the convex hull of a finite set is compact, we can apply the last sentence of the proposition with L′ = Conv(ℝ^{n}, cpct), getting the first statement of the next theorem. Taking cl = c.h.({0} ∪ −) we similarly get the second.

Theorem 14 (Huhn). For every natural number n,

V(Conv(ℝ^{n})) = V(Conv(ℝ^{n}, cpct)),

and

V(Conv(ℝ^{n})_{≥{0}}) = V(Conv(ℝ^{n}, cpct)_{≥{0}}).
□

However, Conv(ℝ^{n}, cpct) is known also to have interesting elementary properties not possessed by Conv(ℝ^{n}). Let us recall that an extremal point of a convex set means a point which is not in the convex hull of any two other points of the set, and the following result ([9, p.276]).

Theorem (Minkowski). Every compact convex subset of ℝ^{n} is the convex hull of its set of extremal points.

Recall also that a lattice L is called join semidistributive if for all x, y_{1}, y_{2} ∈ L one has

x ∨ y_{1} = x ∨ y_{2} ⟹ x ∨ y_{1} = x ∨ (y_{1} ∧ y_{2}).  (13)

Lemma 15 ( = [1, Theorem 3.4], generalizing [5, Theorem 15]). For every positive integer n, Conv(ℝ^{n}, cpct) is join semidistributive.

Proof. Every extremal point of a join x ∨ y must belong to x or to y, since by definition it cannot arise as a convex combination of other points of x ∨ y, hence if x ∨ y_{1} = x ∨ y_{2} as in the hypothesis of (13), extremal points of this set that do not belong to x must belong to y_{1}, and likewise to y_{2}. Thus every extremal point of x ∨ y_{1} = x ∨ y_{2} belongs to x ∪ (y_{1} ∩ y_{2}); hence by Minkowski’s Theorem the convex hull of the latter set, x ∨ (y_{1} ∧ y_{2}), contains the former set. The reverse inclusion is trivial. □

On the other hand, for n ≥ 2, Conv(ℝ^{n}) is not join semidistributive; indeed, the next result will show the failure in these lattices, as n increases, of successively weaker properties, beginning with join-semidistributivity. Following Geyer [8], let us say that a lattice L is n-join semidistributive for a positive integer n if for all x, y_{1}, …, y_{n+1} ∈ L one has

x ∨ y_{1} = ⋯ = x ∨ y_{n+1} ⟹ x ∨ y_{1} = x ∨ (⋁_{i,j; i ≠ j} y_{i} ∧ y_{j}).  (14)

Thus, join semidistributivity is the n = 1 case. One defines n-meet semidistributivity dually.

The (n+4)-element lattice of height 2, M_{n+2}, with least element 0, greatest element 1, and n+2 incomparable elements y_{1}, …, y_{n+2}, is neither n-join semidistributive nor n-meet semidistributive, as may be seen by putting y_{n+2} in the role of x in (14), and in the dual statement. We shall now see that there are several sorts of sublattices with that structure within the lattices Conv(ℝ^{n}). The “open bounded” case of the next result was shown to me by D. Wasserman.

Lemma 16 (Wasserman and Bergman). For every n > 1, Conv(ℝ^{n}) contains copies of M_{n+1} consisting of open bounded sets and copies consisting of closed unbounded sets, in both cases with least element ∅; and Conv(ℝ^{n})_{≥{0}} contains copies of M_{n+1} consisting of bounded sets, with least element {0}. Also, Conv(ℝ^{2})_{≥{0}} (and hence Conv(ℝ^{n})_{≥{0}} for all n > 1) contains copies of M_{c}, the height- 2 lattice of continuum cardinality (and hence contains copies of its sublattices M_{m} for all natural numbers m) consisting of vector subspaces, with least element {0}.

In particular, for n > 1 neither Conv(ℝ^{n}) nor Conv(ℝ^{n})_{≥{0}} is N-join or N-meet semidistributive for any N, and the sublattices of open bounded sets in Conv(ℝ^{n}) and of bounded sets in Conv(ℝ^{n})_{≥{0}} are not (n−1)-join or (n−1)- meet semidistributive.

Proof. To get a copy of M_{n+1} consisting of open bounded sets, start with any n-simplex x, let its faces be x_{1}, …, x_{n+1}, choose an interior point p of x, and for each i let y_{i} be the interior of the n-simplex {p} ∨ x_{i} (or any open convex subset of that n-simplex which has the whole face x_{i} in its closure). We see that the join of any two of the y ’s will have in its closure two faces of x, hence all vertices of x, hence its closure must be x, hence being itself open and convex, it must be the interior of x. On the other hand, the pairwise intersections of the y_{i} are all empty. Hence the lattice generated by these sets is isomorphic to M_{n+1}.

For the closed unbounded example with least element again ∅, extend each of the sets in the preceding example to an infinite cone with apex p, displace each of these cones away from p (say by translating it by the vector from the opposite vertex of x to p), and take their closures. Then every pairwise join is seen to be the whole of ℝ^{n}, while every pairwise meet is again empty.

For the bounded example in Conv(ℝ^{n})_{≥{0}}, take the first example above, assuming p = 0, and use as our new y_{i} the union of the y_{i} of that example with {0}. (Thus, 0 will be the unique boundary-point belonging to each of these sets.)

Finally, an M_{c} in Conv(ℝ^{2})_{≥{0}} is given by the set of all lines through 0. □

There are also cases where we can show the failure of m-join semidistributivity in a natural lattice of convex sets, but where that lattice probably does not contain a copy of M_{m+2}. To get such an example (for any m ≥ 1) in the lattice of convex open subsets of ℝ^{2} containing 0, take m+2 distinct lines through 0 and “thicken” these to open sets x, y_{1}, …, y_{m+1} of width 1. The lattice that these generate will not be M_{m+2}, but clearly fails to satisfy (14). We can also get such examples for bounded open sets in ℝ^{2}, though in this case they fail to have a common point: Fix a triangle T, and let x be the interior of any triangle lying inside T and sharing one edge with T, but not the opposite vertex p. Then take m+1 “small narrow” triangles inside T that have p as a common vertex but no other point in common, and the convex hull of whose union is disjoint from x, and let y_{1}, …, y_{m+1} be their interiors.

The parenthetical comment at the end of the first sentence of the proof of the Lemma 16 shows that the shapes of the convex sets forming a copy of M_{n+1} in the lattice of open bounded convex sets are not unique; but I don’t know an example where the top element of such a sublattice is not an open n-simplex. It would also be of interest to know whether that lattice contains copies of M_{n+2}, and if not, whether it is n-join or n-meet semidistributive. We shall obtain a few related results in later sections.

Jónsson and Rival [13, Lemma 2.1] show that a lattice is join and meet semidistributive if and only if two auxiliary overlattices contain no isomorphic copies of any member of a certain list of 6 lattices, beginning with M_{3}. The above “small narrow triangle” construction gives, when m = 2, a copy of the lattice L_{4} of their list.

Incidentally, Geyer’s concept of n-join semidistributivity, which we have been using, does not have any obvious relationship with Huhn’s n-distributivity. Although for n = 1 they give the conditions of join-semidistributivity and distributivity respectively, of which the latter implies the former, no such implication holds for larger n. For instance, the lattice M_{c} is 2-distributive in Huhn’s sense, but it is not N-join semidistributive for any natural number N in Geyer’s sense.

8. Open bounded sets do not satisfy additional identities

Let us denote by Conv(ℝ^{n}, o.bdd.) the lattice of open bounded convex subsets of ℝ^{n}. We shall show that this lattice, like Conv(ℝ^{n}, cpct), satisfies the same identities as Conv(ℝ^{n}). The idea is that compact sets can be approximated by open bounded sets containing them, from which we shall deduce that any identities of Conv(ℝ^{n}, o.bdd.) are also identities of Conv(ℝ^{n}, cpct), and so by Theorem 14 are identities of Conv(ℝ^{n}).

To approximate compact sets by open sets, we need a different topology on 2^{ℝ^{n}} from the one used earlier; let us again describe this in a general context. If X is any topological space, we may topologize 2^{X} using a basis of open sets with the same form as before,

U(A, B) = {Y ∈ 2^{X} | A ⊆ Y ⊆ B},  (15)

but where, this time, A ranges over all subsets of X, while B is restricted to open subsets. If X is Hausdorff (or even T1), we see that this family of open sets again includes those defining the power-set topology, so our topology is again Hausdorff. Note that for each A ∈ 2^{X}, the sets (15) with A as first argument and with second argument containing A form a neighborhood basis of A in 2^{X}. Thus in proving the next lemma, we shall take it as understood that to be “sufficiently close to” a set A means to contain A and be contained in some specified open neighborhood B of A in X.

In the formulation of that lemma, note that the statement that a function is continuous at arguments with a given property does not simply mean that the restriction of the function to the set of such arguments is continuous, but, more, that such arguments are points of continuity of the whole function.

Lemma 17. If X is any topological space, then the binary operation ∪ : 2^{X}×2^{X} → 2^{X} is continuous in the topology described above; if X is normal (i.e., if disjoint closed subsets of X have disjoint open neighborhoods) then the binary operation ∩ is continuous at arguments given by pairs of closed sets, and if X = ℝ^{n} with the usual topology, then the unary convex-hull operation c.h. : 2^{X} → 2^{X} is continuous at compact sets.

Proof. To show continuity of ∪, consider sets A_{1}, A_{2}, and an open neighborhood B of A_{1} ∪ A_{2} in X. Then we see that the union of any member of U(A_{1}, B) and any member of U(A_{2}, B) contains A_{1} ∪ A_{2} and is contained in B, as required.

For the case of ∩, let A_{1} and A_{2} be closed sets, and B any open neighborhood of A_{1} ∩ A_{2} in X. Then A_{1} − B and A_{2} − B are disjoint closed sets, hence they have disjoint open neighborhoods C_{1} and C_{2}. We see that B ∪C_{1} and B ∪C_{2} will be open neighborhoods of A_{1} and A_{2} which intersect in B (by distributivity of the lattice 2^{X}), and it follows that the intersection of a member of U(A_{1}, B ∪ C_{1}) and a member of U(A_{2}, B ∪ C_{2}) will belong to U(A_{1} ∩ A_{2}, B), as required.

For the final assertion, let A be a compact subset of ℝ^{n}, and B any open neighborhood of c.h.(A). By compactness of c.h.(A), there is some ϵ > 0 such that the set C of all points having distance < ϵ from c.h.(A) is contained in B. This set C is a convex open neighborhood of c.h.(A), hence c.h. carries U(A, C) into U(c.h.(A), C) ⊆ U(c.h.(A), B), as required. □

(I played with several topologies before getting the one that made the above result – in particular, continuity of intersection – easy to prove. Some of these might be preferable for other considerations of the same sort. Under the above topology, every open set A is an isolated point, since U(A, A) is a singleton. If one wants to approximate open sets by larger open sets, one might prefer a weaker topology in which the conditions on Y in (15) are, say, strengthened to A ⊆ Y, cl(Y) ⊆ B.)

It follows from Lemma 17 that in the topology we have defined, the lattice operations of Conv(ℝ^{n}) are continuous at arguments belonging to Conv(ℝ^{n}, cpct). Moreover, Conv(ℝ^{n}, cpct) lies in the closure of Conv(ℝ^{n}, o.bdd.), since every compact convex set A is the limit in this topology, as ϵ → 0, of the open convex set of points at distance < ϵ from A. Hence any lattice identities holding in Conv(ℝ^{n}, o.bdd.) also hold in Conv(ℝ^{n}, cpct). The same considerations apply to the pair of lattices Conv(ℝ^{n}, o.bdd.)_{≥{0}} and Conv(ℝ^{n}, cpct)_{≥{0}}. In view of Theorem 14, these observations give us Theorem 18. For every natural number n, V(Conv(ℝ^{n}))=V(Conv(ℝ^{n}, o.bdd.)), and V(Conv(ℝ^{n})_{≥{0}}) = V(Conv(ℝ^{n}, o.bdd.)_{≥{0}}). □

Let us note that there is an order-preserving bijection between the elements of Conv(ℝ^{n}, o.bdd.) and those compact convex subsets of ℝ^{n} which, if nonempty, have nonempty interior, given by the operation of topological closure, with inverse given by topological interior. This is not, however, an isomorphism between sublattices of Conv(ℝ^{n}), because the second of these sets is not closed under intersection. (E.g., consider two adjacent closed polygons in ℝ^{2}. Nor can we get around this problem by going to a homomorphic image of Conv(ℝ^{n}, cpct) where sets with empty interior are identified with ∅, since the join of two such sets can have nonempty interior.)

On the other hand, the set of compact convex sets which are neighborhoods of 0 is a sublattice of Conv(ℝ^{n}, cpct)_{≥{0}}, and the above correspondence gives us an isomorphism between it and Conv(ℝ^{n}, o.bdd.)_{≥{0}}; so properties of Conv(ℝ^{n}, cpct)_{≥{0}} which carry over to sublattices also hold for Conv(ℝ^{n}, o.bdd.)_{≥{0}}. (More generally, any sublattice of Conv(ℝ^{n}, o.bdd.) whose members have a common point p is contained in Conv(ℝ^{n}, o.bdd.)_{≥{p}} ≅ Conv(ℝ^{n}, o.bdd.)_{≥{0}}, and so can be studied in the same fashion.) Thus, despite the examples of Lemma 16, we have

Corollary 19 (to Lemma 15). Conv(ℝ^{n}, o.bdd.)_{≥{0}} is join semidistributive. Hence, every counterexample to join semidistributivity in Conv(ℝ^{n}, o.bdd.) has the property that the intersection of the three sets involved is empty. □

Since we are considering elementary properties in which the lattices Conv(ℝ^{n}) and Conv(ℝ^{n}, o.bdd.) agree or differ, we should note the obvious difference, that the former is atomistic (every element is a possibly infinite join of atoms), while the latter has no atoms. Cf. [2] and papers referred to there, in which lattices of convex sets and related structures are characterized in terms of properties of their atoms, and also [1].

We noted above that lattices Conv(ℝ^{n})_{≥S} for nonempty S satisfy all identities holding in Conv(ℝ^{n})_{≥{0}}. Let us end this section by using Theorem 18 to show that for bounded S, the converse is also true.

Corollary 20 (to Theorem 18 and proof of Theorem 14). For every natural number n and every bounded set S ⊆ ℝ^{n},

V(Conv(ℝ^{n})_{≥S}) = V(Conv(ℝ^{n}, cpct)_{≥S})
= V(Conv(ℝ^{n}, o.bdd.)_{≥S}) = V(Conv(ℝ^{n})_{≥{0}}).

Proof. By a translation, we can assume without loss of generality that 0 ∈ S; thus the last of the above lattices contains all the others, so letting f(x_{1}, …, x_{m}) = g(x_{1}, …, x_{m}) be any identity not satisfied there, it suffices to prove that f = g is not satisfied identically in any of the other lattices.

Now Theorem 18 shows that f = g is not an identity of Conv(ℝ^{n}, o.bdd.)_{≥{0}}, hence we can choose open bounded convex sets x_{1}, …, x_{m} containing 0 which do not satisfy it. The intersection of these sets is a neighborhood of the origin, and dilating the x_{i} by a large enough real constant, we can assume without loss of generality that this neighborhood contains S. Hence f = g is also not an identity of Conv(ℝ^{n}, o.bdd.)_{≥S}, hence not an identity of Conv(ℝ^{n})_{≥S} either.

The case of Conv(ℝ^{n}, cpct)_{≥S} is similar. Again take elements x_{1}, …, x_{m} ∈ Conv(ℝ^{n}, o.bdd.)_{≥{0}} not satisfying f = g. We saw in the proof of Theorem 14 that if we approximate x_{1}, …, x_{m} closely enough from below, in the topology of that proof, by compact convex subsets y_{1}, …, y_{m}, then these approximating sets will also fail to satisfy that identity. Since the intersection of the x_{i} is a neighborhood of 0, it contains an n-simplex with 0 in its interior; so we can take all the y_{i} to contain the finitely many vertices of that simplex, hence to be neighborhoods of 0. As before, we may now dilate them so that they all contain S, getting the required elements of Conv(ℝ^{n}, cpct)_{≥S}. □

9. The possibility of surface phenomena

The technique by which we just proved Corollary 20 can be inverted to show that if T ⊆ ℝ^{n} is any convex set with nonempty interior, then the lattice of convex sets contained in T, and its sublattices of compact convex subsets of T and open bounded convex subsets of T, satisfy the same identities as Conv(ℝ^{n}). Namely, given any identity not holding in Conv(ℝ^{n}), we already know that we can find elements x_{1}, …, x_{m} not satisfying it in Conv(ℝ^{n}, cpct). Assuming without loss of generality that 0 lies in the interior of T, we can shrink x_{1}, …, x_{m} by a constant so that they are all contained in T; likewise we can, as before, approximate compact convex sets by open bounded convex sets. A similar argument shows that the lattice of convex subsets of T which contain a specified point of the interior of T satisfies the same identities as Conv(ℝ^{n})_{≥{0}}.

However, if we specify two convex sets S ⊆ T, say with S compact and T open, and look at the interval [S, T] = {x ∈ Conv(ℝ^{n}) | S ⊆ x ⊆ T }, it is not clear whether, for some choices of S and T, this may satisfy more identities than hold in Conv(ℝ^{n})_{≥{0}}. (Picturing S and T as “very close”, e.g., the closed ball of radius 1 and the open ball of radius 1 + ϵ, explains, I hope, the title of this section.)

Let us relax our assumptions on S and T for a moment and look at a more extreme example. If we take for S the open unit ball and for T the closed unit ball in ℝ^{n}, then every set x with S ⊆ x ⊆ T is convex, so in this case [S, T] may be identified with the lattice of all subsets of the unit sphere, which is distributive, though we have seen that Conv(ℝ^{n})_{≥{0}} is not even (n−1)-distributive.

In the case where S is compact and T open and nonempty, however, things cannot go that far:

Lemma 21. Let S ⊆ T be convex subsets of ℝ^{n}. If S is compact and T is open and nonempty, or more generally, if some hyperplane P ⊆ ℝ^{n} disjoint from S intersects T in a set with nonempty relative interior (i.e., is such that P ∩ T is (n−1)-dimensional), then V([S, T]) ⊇ V(Conv(ℝ^{n−1})).

Proof. Clearly, a pair S ⊆ T satisfying the first hypothesis satisfies the second, so let us assume the latter. Let H be the closed half-space in ℝ^{n} bounded by P that contains S (or if S is empty, either of the closed half-spaces bounded by P), and consider the sublattice [S, H ∩ T] ⊆ [S, T]. The operation of intersecting with P can be seen to give a lattice homomorphism [S, H ∩ T] → [∅, P ∩ T], and this is surjective, since it has the set-theoretic section x ↦ S ∨ x. Since P ∩ T has nonempty interior in P, the observations of the first paragraph of this section show that [∅, P ∩ T] satisfies precisely the identities of Conv(ℝ^{n−1}). Hence [S, H ∩ T], since it maps homomorphically onto [∅, P ∩ T], cannot satisfy any identities not satisfied by Conv(ℝ^{n−1}), so neither can the larger lattice [S, T]. □

It is not evident whether, for S compact and T open, V([S, T]) can ever be strictly smaller than V(Conv(ℝ^{n})_{≥{0}}); nor, for that matter, whether it can ever fail to be strictly smaller, if S has nonempty interior and T is bounded. Another sort of interval [S, T] ⊆ Conv(ℝ^{n}) whose identities it would be interesting to investigate is given by letting S = {0} and T be a closed half-space with 0 on its boundary. Again these identities must lie somewhere between those of Conv(ℝ^{n−1}) and Conv(ℝ^{n})_{≥{0}}.

In this and preceding sections we have used from time to time the fact that translations and dilations preserve convexity. More generally, if φ is any projective transformation on n-dimensional projective space ℙ^{n} ⊇ ℝ^{n}, then convex subsets of ℝ^{n} which do not meet the hyperplane that φ sends to infinity are taken by φ to convex sets. Hence if S ⊆ T are two such convex sets, φ induces a lattice isomorphism [S, T] ≅ [φ(S), φ(T)]. This observation might be useful in classifying the varieties determined by such intervals.

Still another class of sublattices of Conv(ℝ^{n}) one might investigate is that of all convex sets that are carried into themselves by a given affine map; e.g., the orthogonal projection of ℝ^{n} onto a specified subspace. (If that subspace is {0}, we get Conv(ℝ^{n})_{≥{0}} with one additional element ∅ thrown in.)

10. Dualities

Let us recall the definition of a concept we have referred to a couple of times in passing. A closed half-space in ℝ^{n} means a set of the form

{p ∈ ℝ^{n} | f(p) ≤ λ},  (16)

for some nonzero linear functional f on ℝ^{n} and some real number λ. It is a standard result that every closed convex subset of ℝ^{n} is an intersection of closed half-spaces.

The half-space (16) contains the point 0 if and only if λ is nonnegative, hence closed convex sets containing 0 can be characterized as intersections of half-spaces (16) having λ ≥ 0. Such sets can, in fact, be expressed as intersections of such half-spaces with λ > 0, since a half-space (16) with λ = 0 is the intersection of all the half-spaces with the same f and positive λ. But a half-space (16) with λ positive can be written as {p ∈ ℝ^{n} | λ^{−1}f(p) ≤ 1}, or, expressing the linear functional λ^{−1}f as the dot product with some q ∈ ℝ^{n}, as

{p ∈ ℝ^{n} | q·p ≤ 1}.  (17)

Thus, for any subset S ⊆ ℝ^{n}, if we define

S^{∗} = {q ∈ ℝ^{n} | (∀p ∈ S) q·p ≤ 1},  (18)

then S^{∗∗} will be the least closed convex set containing S ∪ {0}. Moreover, we see that S^{∗} will also be a closed convex subset containing 0, which uniquely determines and is determined by S^{∗∗}. Thus, the operator ^{∗} gives a bijection of the family of all closed convex sets containing 0 with itself, which is easily seen to be inclusionreversing. (This is an example of a Galois connection; cf. [3, §5.5] for a general development of the concept, with many examples.) Let us call two closed convex sets containing 0 that are related in this way dual to one another. (The dual of a convex set is sometimes called its polar set, e.g., in [15].) Examples in ℝ^{3} are a cube and an octahedron of appropriate radii centered at the origin, and similarly a dodecahedron and an icosahedron. The unit sphere is self-dual.

The class of closed convex subsets of ℝ^{n} forms a lattice (by general properties of Galois connections), which, like Conv(ℝ^{n})_{≥{0}}, has intersection as its meet operation; but the join operations do not everywhere coincide – a consequence of the fact that for closed sets u, v ∈ Conv(ℝ^{n})_{≥{0}}, their join in that lattice, c.h.(u ∪ v), may not be closed, so that to get their join as closed convex sets, one must take its topological closure. For example, let n = 2, and let u be the closed strip {(x, y) | −1 ≤ y ≤ 1} and v the line segment c.h.(0, (0, 2)). Then the join of u and v in Conv(ℝ^{n})_{≥0} is

{(x, y) | −1 ≤ y < 2} ∪ {(0, 2)},

while their join in the corresponding lattice of closed convex sets is {(x, y) | −1 ≤ y ≤ 2}. In view of this difference in operations, care is needed when using our duality on closed convex sets to deduce results about the lattice Conv(ℝ^{n})_{≥{0}}.

If x and y are mutually dual closed convex sets containing 0 in ℝ^{n}, it is not hard to see that one of them is bounded (i.e., compact) if and only if 0 is an interior point of the other. It follows that the class of closed bounded convex sets having 0 in their interior is self-dual; moreover, as we noted at the beginning of §7, the join in Conv(ℝ^{n}) of two compact sets is again compact, from which it follows that unlike the lattice of all closed convex sets containing 0, this is a sublattice of Conv(ℝ^{n}). It is easy to see from the equality of the second and fourth varieties in Corollary 20 that this sublattice satisfies the same identities as Conv(ℝ^{n}). Hence the existence of the anti-automorphism just noted gives us

Theorem 22. The class of lattice identities satisfied by Conv(ℝ^{n})_{≥{0}} is self-dual, i.e., closed under interchanging all instances of ∨ and ∧. Equivalently, the variety V(Conv(ℝ^{n})_{≥{0}}) is closed under taking dual lattices. □

And indeed, the identities proved for Conv(ℝ^{n})_{≥{0}} in Theorems 8 and 11 respectively are dual to one another. This is not true of the identities proved for Conv(ℝ^{n}) in those theorems. In fact, it is easy to verify

Exercise 23. (i) For every positive integer n, show by example that Conv(ℝ^{n}) does not satisfy either of the identities

((D_{n}^{op}) ∧ z) ∨ y_{1} ∨ y_{2} and (((D_{n}) ∨ z′) ∧ z) ∨ (⋁_{i} y_{i}),

dual to the identities proved for that lattice in Theorems 8 and 11 respectively.

(ii) Show that these observations together with Theorem 22 yield an alternative way of verifying that in each of Theorems 8 and 11, the identities proved for Conv(ℝ^{n}) are not satisfied by Conv(ℝ^{n+1})_{≥{0}}.

Theorem 22 also implies that Conv(ℝ^{n+1})_{≥{0}}, and hence also Conv(ℝ^{n}), satisfies the dual of the identity of Theorem 12, which we had not previously obtained.

Just as every closed convex set is an intersection of closed half-spaces (16), so every open convex set is an intersection of open half-spaces,

{p ∈ ℝ^{n} | f(p) < λ}.

But here the converse is not true. Indeed, every closed half-space is also an intersection of open half-spaces, so the class of intersections of open half-spaces includes both the open and the closed convex sets. If for every subset S ⊆ ℝ^{n} one defines

S^{⋄} = {q ∈ ℝ^{n} | (∀p ∈ S) q·p < 1},  (19)

one gets a duality theory for the class of ^{⋄}^{⋄}-invariant sets, i.e., sets satisfying x^{⋄}^{⋄} = x, a larger class than that covered by the preceding duality. It is not hard to see that a necessary and sufficient condition for a set x ∈ Conv(ℝ^{n})_{≥{0}} to be ^{⋄⋄}-invariant is that for every point p of the boundary of x which does not belong to x, there exist a supporting hyperplane of x through p containing no point of x. From this we can see that the union of the open unit ball B in ℝ^{n} with any subset of its boundary is ^{⋄}^{⋄}-invariant; its ^{⋄}-dual is the union of B with the complementary subset of the boundary. On the other hand, we find that the union of an open polygonal neighborhood P of 0 in the plane with a nonempty finite (or countable) subset S of its boundary is never ^{⋄}^{⋄}-invariant, since by the above characterization of ^{⋄}^{⋄}-invariant sets, each point q ∈ S forces the ^{⋄}^{⋄}-closure of P ∪ S to contain all points p of the open edge(s) of our polygon containing or adjacent to q.

Like the ^{∗∗}-invariant sets (closed convex sets containing 0), the ^{⋄}^{⋄}-invariant subsets of ℝ^{n} form a lattice by general properties of Galois connections, which has the same meet operation as Conv(ℝ^{n})_{≥{0}}, but different join operation. For instance, if P is, as above, an open polygonal neighborhood of 0 in ℝ^{2}, q a point of its boundary, and Q = c.h.(0, q), then writing ∨ for the join operation of Conv(ℝ^{2})_{≥{0}}, we find that P ∨ Q = P ∪ Q = P ∪ {q}, but as we saw above, this set is not ^{⋄}^{⋄}-invariant. Rather, (P ∪ Q)^{⋄}^{⋄}, the join of P and Q in the lattice of ⋄^{⋄}-invariant sets, is obtained by attaching to P ∪ {q} the open edge(s) containing or adjacent to q. Further, the ^{∗∗}-invariant subsets of ℝ^{n} do not form a sublattice of the the ^{⋄}^{⋄}-invariant sets. To see this in ℝ^{2}, let

u = {(x, y) | |x| < 1, y ≥ 1/(1 − x^{2}) − 2},
v = {(x, y) | |x| < 1, y ≤ −1/(1 − x^{2}) + 2}.

We see that the join of u and v in Conv(ℝ^{2})_{≥{0}} is {(x, y) | |x| < 1}, which is open, hence is ^{⋄}^{⋄}-invariant, hence is their join in the lattice of ^{⋄}^{⋄}-invariant sets; but it is not closed, hence is not their join in the lattice of ^{∗∗}-invariant sets.

But again, one can apply ^{⋄}^{⋄}-duality to sublattices of Conv(ℝ^{n})_{≥{0}} which consist of ^{⋄}^{⋄}-invariant elements, since for these, the two lattice structures in question must agree. In particular, one can verify that ^{⋄}-duality interchanges compact and open sets, giving an anti-isomorphism between the sublattices Conv(ℝ^{n}, cpct)_{≥{0}} and Conv(ℝ^{n}, open)_{≥{0}}. (This anti-isomorphism can also be obtained by composing the anti-isomorphism “^{∗}” between Conv(ℝ^{n}, cpct)_{≥{0}} and the lattice of closed convex sets that are neighborhoods of the origin, with the isomorphism noted earlier between that lattice and Conv(ℝ^{n}, open)_{≥{0}}.) The existence of this antiisomorphism gives

Corollary 24 (to Lemma 15). For every positive integer n, Conv(ℝ^{n}, open)_{≥{0}} is meet semidistributive, that is, satisfies the dual of (13). □

This explains the fact that in Lemma 16 and the discussion that followed, though we saw that Conv(ℝ^{n}, open)_{≥{0}} was not join semidistributive, we found no copies of M_{3} in it.

Meet-semidistributivity does not hold in any of the lattices of convex sets we have considered that are not defined so as to make those sets all have some point (such as 0) in common; for in each of these lattices, it is easy to get examples of a set x having empty intersection with each of two sets y_{1} and y_{2}, but nonempty intersection with their join. It also does not hold in Conv(ℝ^{n}, cpct)_{≥{0}} for n > 1, since that lattice contains an embedded copy of Conv(ℝ^{n−1}, cpct) (cf. proof of Lemma 1), to which the above observation applies.

While on the topic of join- and meet-semidistributivity, I will note some questions and examples which are easier to state now that we have named several sublattices of Conv(ℝ^{n}). Kira Adaricheva (personal communication) has posed several questions of the following form: If we take one of the sublattices of Conv(ℝ^{n}) that we know to be join- or meet-semidistributive, and extend it by adjoining, within Conv(ℝ^{n}), a single “nice” outside element, is the property in question already lost, and if so, does the resulting lattice in fact contain copies of M_{3}?

In many cases this does happen. For instance, the sublattice L ⊆ Conv(ℝ^{2}) generated by the join-semidistributive sublattice Conv(ℝ^{2}, cpct), and the open disk, which we shall denote B, contains a copy of M_{3}. To describe it, let p_{0}, p_{1}, p_{2} be any three distinct points on the unit circle, and for i = 0, 1, 2 define the point q_{i} = 1/5 p_{i} + 2/5 p_{i+1} + 2/5 p_{i+2} (subscripts evaluated mod 3; I suggest making a sketch). Let x_{i} = (c.h.(q_{i}, p_{i+1}) ∧ B) ∨ (c.h.(q_{i}, p_{i+2}) ∧ B). Then x_{1}, x_{2}, x_{3} can be seen to generate a copy of M_{3} in Conv(ℝ^{2}), resembling, though not identical to, the n = 2 case of the first example described in Lemma 16. An analogous construction gives a family of elements of L resembling the example described immediately following that lemma, showing that L is not m-join semidistributive for any m.

As another example let L be the sublattice of Conv(ℝ^{3})_{≥{0}} generated by Conv(ℝ^{3}, o.bdd.)_{≥{0}} and the cube C = [−1, +1]^{3}. Letting B denote the interior of C, I claim that for any triangle T drawn on a face F of C, L contains the union of B with the interior of T relative to F. Indeed, we can find in Conv(ℝ^{3}, o.bdd.)_{≥{0}} an open pyramid P (with apex near 0) meeting no face of C except F, and meeting F in precisely the relative interior of T. Then (P ∧ C) ∨ B will be the desired set. If we construct three sets of this sort using as our T ’s three triangles in a common face F of C, whose relative interiors form a copy of the n = 2 case of the first example of Lemma 16, then the sublattice generated by the three resulting sets (P ∧ C) ∨ B will have the form M_{3} (with B as its least element; cf. the last sentence of Corollary 19). The same trick can again be used to get examples like those of the paragraph following Lemma 16, so this L, too, is not m-join semidistributive for any m.

On the other hand, I do not know what can be said about the lattices gotten by adjoining to Conv(ℝ^{n}, o.bdd.)_{≥{0}} a closed ball D about 0, respectively an open ball B not containing 0, except that the second of these lattices is not meetsemidistributive, since we can find three “fingers” in Conv(ℝ^{n}, o.bdd.)_{≥{0}} which meet B in disjoint subsets x, y_{1}, y_{2}, such that x ∧ (y_{1} ∨ y_{2}) is nonempty.

11. Relatively convex sets

Let S be a subset of ℝ^{n}, in general non-convex. We shall call a subset x ⊆ S convex relative to S if x = c.h.(x) ∩ S; equivalently, if x is the intersection of S with some convex subset of ℝ^{n}. Such sets x will form a lattice RelConv(S), with meet operation given, as in (2), by intersection, but join now given by

x ∨ y = c.h.(x ∪ y) ∩ S.  (20)

Note that the map x ↦ c.h.(x) gives a bijection between the relatively convex subsets of S, and the subsets of ℝ^{n} which are convex hulls of subsets of S, the inverse map being given by −∩S. Convex sets of the latter sort form a lattice with join as in (2), but with meet operation

x ∧ y = c.h.(x ∩ y ∩ S).  (21)

We observe that Carathéodory’s Theorem (with the “refinement” given in the final sentence), regarded as a statement about the closure operator c.h.(−) on ℝ^{n}, entails the same properties for the closure operator c.h.(−) ∩ S on S. Hence the proofs of Lemma 3 and of the positive assertions of Theorem 8 immediately yield

Proposition 25 (cf. Huhn [11, Lemma 3.2]). If n is a natural number and S a subset of ℝ^{n}, then RelConv(S) satisfies the identity ((D_{n}) ∨ z) ∧ y_{1} ∧ y_{2}, and if p is a point of S, RelConv(S)_{≥{p}} satisfies the identity D_{n}. □

However, the next lemma shows that the corresponding results fail badly for the identities involving D_{n}^{op} obtained in Theorem 11. In this lemma, the second assertion embraces the first (plus two obvious intermediate results not stated); however I include the first assertion because both its statement and the example proving it are more transparent than for the second.

Lemma 26. For S a subset of ℝ^{2}, RelConv(S) need not satisfy the identity D_{n}^{op} for any positive integer n. In fact, for p an element of such an S, RelConv(S)_{≥{p}} need not satisfy the identity (((D_{n}^{op}) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}).

Proof. To get the first assertion, let S be the set consisting of the unit circle and its center, 0. Given n, let q_{1}, …, q_{2n+2} be the successive vertices of a regular (2n+2)-gon on that circle; for i = 1, …, n+1, let y_{i} = {q_{1}, …, q_{n+1}} − {q_{i}}, and let x = {q_{n+2}, …, q_{2n+2}}. Observe that each intersection of n of the y ’s consists of one of the points q_{i}, and that x contains the antipodal point; hence the join of x with each intersection of n y ’s contains 0. On the other hand, the intersection of all the y ’s is empty, hence its join with x is x, which does not contain 0. So D_{n}^{op} fails for this choice of arguments.

To get an example where all the given sets have a common element p, and where, moreover, the indicated weaker identity fails, let us take for S the same set as in the preceding example, with the addition of one arbitrary point r outside the circle, at distance ≥ 2 from 0. This time, let q_{1}, …, q_{2n+6} be the successive vertices of a regular (2n+6)-gon on S, placed so that q_{n+2} lies on the line connecting 0 with the external point r. For i = 1, …, n+1, let y_{i} = {q_{1}, …, q_{n+3}} − {q_{i}}, and let x = {q_{n+3}, …, q_{2n+4}}. We note that the set of subscripts of the q ’s occurring in each of these sets lies in an interval of length < n+3, so the absence of the point 0 does not contradict convexity of these sets relative to S. Taking p = q_{n+3} we see that all of these sets belong to RelConv(S)_{≥{p}}.

As in the previous example, the intersection of any n of the y ’s contains one of q_{1}, …, q_{n+1}, and x contains its antipodal point, so that the join of x with that intersection contains 0, hence so does the intersection of all these joins, i.e., the value of the right-hand side of D_{n}^{op} at these arguments. On the other hand, the intersection of all the y ’s is {q_{n+2}, q_{n+3}}, and the union of this set with x still has all subscripts lying in an interval of length < n+3, so the join of those two sets, the left-hand side of D_{n}^{op}, does not contain 0. Hence if we take z′ = {q_{n+3}, 0}, the intersection of the right-hand side of D_{n}^{op} with z′ is z′ = {q_{n+3}, 0}, while intersection of the left-hand side with z′ is {q_{n+3}}. If, finally, we let z = {q_{n+3}, r} and take the joins of this element with those two intersections, we see that in the first case the resulting set contains the point q_{n+2}, since we assumed this to lie on the line-segment from 0 to r, while in the second, it does not. Since q_{n+2} ∈ ⋀_{i} y_{i}, the two sets remain distinct on intersecting with ⋀_{i} y_{i}, showing the failure of (((D_{n}^{op}) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}). □

The above example in ℝ^{2} is also an example in ℝ^{n} for any n ≥ 2. For completeness, we should also consider dimensions n = 0 and 1. If we look at the chain of varieties corresponding to that of Lemma 1, but with each V(Conv(ℝ^{n})) replaced by the variety generated by all the lattices RelConv(S) for subsets S ⊆ ℝ^{n}, and each V(Conv(ℝ^{n})_{≥{0}}) by the variety generated by lattices RelConv(S)_{≥{p}}, then we see that the first term of this chain is still the trivial variety and the next two still satisfy the distributive identity (in the last case, by the final assertion of Proposition 25 for n = 1). Since the distributive identity implies the identities of every nontrivial variety, we conclude that allowing lattices of relatively convex sets has not enlarged the varieties we get at these three steps. We have just shown the contrary from the fifth step on; this leaves only the fourth step, i.e., the relation between V(Conv(ℝ^{1})) and the variety generated by all lattices of form RelConv(S) with S ⊆ ℝ^{1}. Here again, it turns out that we have equality. This follows from our observation that RelConv(S) is isomorphic to the lattice of convex hulls of subsets of S, together with

Lemma 27. Let S be a subset of ℝ^{1}. Then the convex hulls of subsets of S form a sublattice of Conv(ℝ^{1}).

Proof. We have noted, for arbitrary n and S ⊆ ℝ^{n}, that the convex hulls of subsets of S are closed under the join operation of the full lattices of convex sets, so we need only show them closed under meets, i.e., intersections, when n = 1. This comes down to showing that if p ∈ c.h.(x) ∩ c.h.(y), where x and y are relatively convex subsets of S, then p ∈ c.h.(x ∩ y). The fact that p ∈ c.h.(x) means that q_{1} ≤ p ≤ q_{2} for some q_{1}, q_{2} ∈ x; similarly, q_{3} ≤ p ≤ q_{4} for some q_{3}, q_{4} ∈ y. From the relative convexity of x and y and the order-relations of these elements, we now see that max(q_{1}, q_{3}) ∈ x ∩ y and min(q_{2}, q_{4}) ∈ x ∩ y. Hence p ∈ c.h.(max(q_{1}, q_{3}), min(q_{2}, q_{4})) ⊆ c.h.(x ∩ y). □

For further results on V(Conv(ℝ^{1})) and its subvarieties, see [17].

Incidentally, the lattice RelConv(S), for S the set used in the first part of the proof of Lemma 26 (consisting of the unit circle and its center) shows that none of the identities (((D_{m}^{op}) ∧ z′) ∨ z) ∧ (⋀_{i} y_{i}) implies any of the identities D_{n}^{op}. Indeed, we showed that RelConv(S) satisfies none of the latter identities; let us now show that (unlike the lattice considered in the second part of that proof) it satisfies all of the former. It suffices to verify the strongest of these, the case m = 1. Writing S = C ∪ {0}, where C is the unit circle, we see that the map − ∩ C : RelConv(S) → RelConv(C) = 2^{C} is a homomorphism; so if two lattice expressions in x, y_{1}, y_{2}, z, z′ are identically equal in distributive lattices, their values in RelConv(S) will always agree except, perhaps, as to whether they contain 0. We now consider separately the cases 0 ∈ y_{1} ∧ y_{2} and 0 ∉ y_{1} ∧ y_{2}. In the former case, the two sides of D_{1}^{op}(x, y_{1}, y_{2}) agree in containing 0, hence are equal, so a fortiori the two sides of (((D_{1}^{op}) ∧ z′) ∨ z) ∧ (y_{1} ∧ y_{2}) are equal. In the latter case, neither side of the latter relation can contain 0, hence again they are equal.

12. The snowflake

Let us look at a particularly neat example of a lattice of relatively convex subsets of a set S.

Let p_{1}, p_{2}, p_{3}, −p_{1}, −p_{2}, −p_{3} be the successive vertices of a regular hexagon in ℝ^{2} centered at 0. Let

S_{1} = c.h.(p_{1}, −p_{1}), S_{2} = c.h.(p_{2}, −p_{2}), S_{3} = c.h.(p_{3}, −p_{3}),
S = S_{1} ∪ S_{2} ∪ S_{3},

and let L be the sublattice of RelConv(S)_{≥{0}} generated by the three line-segments S_{1}, S_{2} and S_{3}. (In view of the form of S, I think of this example as “the snowflake”.)

Every element of L will clearly be centrally symmetric and topologically closed; hence every such element has the form

λ_{1} S_{1} ∪ λ_{2} S_{2} ∪ λ_{3} S_{3} (λ_{1}, λ_{2}, λ_{3} ∈ [0, 1]).  (22)

The join of S_{1} and S_{2} in this lattice must have the form S_{1} ∪ S_{2} ∪ λ S_{3} for some λ ∈ (0, 1). (Elementary geometry shows that λ = 1/2; but we don’t need to know this now, and will get it from a general formula soon.) Intersecting this join with S_{3}, we see that λ S_{3} ∈ L. By symmetry we also have λ S_{1}, λ S_{2} ∈ L, and we see that these together generate a proper sublattice of L isomorphic to the whole lattice. In particular, L has infinite descending chains of elements, e.g., S_{1} > λ S_{1} > λ^{2}S_{1} > ….

Let’s figure out how to calculate in L. The first thing we should find are the conditions on λ_{1}, λ_{2} and λ_{3} for a set (22) to be relatively convex. Calculation shows that for λ_{1}, λ_{2} and λ_{3} nonzero, the points λ_{1} p_{1}, λ_{2} p_{2} and λ_{3} p_{3} are collinear if and only if λ_{2}^{−1} = λ_{1}^{−1} + λ_{3}^{−1}. (In verifying this, the key relation is p_{2} = p_{1} + p_{3}.) Hence we see that one necessary condition for the convexity of (22) is λ_{2}^{−1} ≤ λ_{1}^{−1} + λ_{3}^{−1}; by symmetry, the remaining conditions are λ_{1}^{−1} ≤ λ_{2}^{−1} + λ_{3}^{−1} and λ_{3}^{−1} ≤ λ_{1}^{−1} +λ_{2}^{−1}. It is easy to verify that if the λ_{i} are also allowed to be zero, and we write 0^{−1} = ∞ and consider ∞ greater than all real numbers, then these three conditions continue to be necessary and sufficient for convexity.

Let us therefore index elements (22) by the three parameters λ_{1}^{−1}, λ_{2}^{−1}, λ_{3}^{−1}, defining

[a_{1}, a_{2}, a_{3}] = a_{1}^{−1}S_{1} ∪ a_{2}^{−1}S_{2} ∪ a_{3}^{−1}S_{3} (a_{1}, a_{2}, a_{3} ∈ [1, ∞]),  (23)

so that the lattice of topologically closed centrally symmetric elements of RelConv(S)_{≥{0}} consists of the sets [a_{1}, a_{2}, a_{3}] with

a_{1} ≤ a_{2} + a_{3},  a_{2} ≤ a_{1} + a_{3},  a_{3} ≤ a_{1} + a_{2}.  (24)

Note that the ordering of this lattice is by reverse componentwise comparison of expressions [a_{1}, a_{2}, a_{3}], and that S_{1}, S_{2}, S_{3} are the elements [1, ∞, ∞], [∞, 1, ∞], [∞, ∞, 1]. Lattice-theoretic meet is clearly given by componentwise supremum, while the lattice-theoretic join of two elements is gotten by first taking their componentwise infimum, which represents their set-theoretic union, then getting its relative convex hull by reducing the largest entry to the sum of the other two if it exceeds this. So, for instance, S_{1} ∨S_{2} = [1, ∞, ∞] ∨ [∞, 1, ∞] is gotten by forming the componentwise infimum, [1, 1, ∞], and then decreasing the last component to the sum of the first two, getting [1, 1, 2] (confirming the value λ = 1/2 in our earlier description of this element). We can now calculate, e.g., the meet (componentwise supremum) of this with S_{3} = [∞, ∞, 1], namely [∞, ∞, 2]; and take the join of this meet with S_{2} = [∞, 1, ∞] by forming the componentwise infimum [∞, 1, 2], and adjusting the first component as above, getting [3, 1, 2].

Note that the only way these lattice operations yield components in their values that did not occur as components in their arguments is by addition; hence, as L was defined to be generated by S_{1}, S_{2} and S_{3}, all finite components a_{i} that occur in the expressions (23) for elements of L are positive integers. It is not hard to verify that all positive integers indeed occur, and that the elements of L are all the elements (23) satisfying (24) with a_{1}, a_{2}, a_{3} ∈ {1, 2, 3, …, ∞}. So we have gotten a very arithmetic description of this lattice.

Though we have seen that L contains an infinite descending chain, it is interesting to note that the sublattice generated by any finite set of elements (23), none of which have any infinite components, is finite; for the lattice operations will not produce, in any position, entries larger than the corresponding entries of their arguments.

It would be of interest to examine lattices of the form RelConv(S)_{≥{0}} for more general sets S which are unions of finitely many line-segments (or rays) through the origin in ℝ^{n}. In this situation, the conditions for convexity are always given by linear inequalities in the “ λ_{i}^{−1}”; let me sketch why.

First, some general observations. Suppose p_{1}, …, p_{m} (m ≥ 3) are a minimal linearly dependent family of vectors in ℝ^{n}; thus they satisfy a nontrivial linear relation ∑ c_{i} p_{i} = 0, unique up to scalars. Given positive real numbers λ_{1}, …, λ_{m}, how can we tell whether the points λ_{1} p_{1}, …, λ_{m} p_{m} lie in an (m−1)-dimensional affine subspace of their m-dimensional span? This will hold if and only if the linear relation satisfied by these modified elements is an affine relation; i.e., has coefficients summing to 0. Those coefficients are c_{i} λ_{i}^{−1}, so the condition is

∑ c_{i} λ_{i}^{−1} = 0.

In this situation, is the affine relation among the λ_{i} p_{i} equivalent to a relation expressing one of them, say λ_{j} p_{j}, as a convex linear combination of the others? One sees that this is so if and only if the coefficient c_{j} is opposite in sign to all the other c_{i}; in that situation, let us rewrite the expression satisfied by the λ_{i}^{−1} as

λ_{j}^{−1} = −c_{j}^{−1} ∑_{i≠j} c_{i} λ_{i}^{−1}.

Note that the abovementioned condition on signs can be satisfied by at most one j, and can be looked at as saying that the ray determined by the corresponding vector p_{j} is in the convex hull of the rays determined by the other p_{i}. When it is satisfied, one finds that a union of intervals ⋃_{i} [0, λ_{i} p_{i}] is relatively convex in the union of the rays determined by the p_{i} if and only if λ_{j} has at least the value given by the above formula, i.e., if and only if

λ_{j}^{−1} ≤ −c_{j}^{−1} ∑_{i≠j} c_{i} λ_{i}^{−1}.  (25)

If the c_{i} do not consist of one of one sign and the rest of the opposite sign, then none of the rays determined by the p_{i} is in the convex hull of the rest, and every set ⋃_{i} [0, λ_{i} p_{i}] (λ_{1}, …, λ_{m} > 0) is relatively convex. (An example of this situation is given for n = 3 by letting p_{1}, …, p_{4} be the vertices of a convex quadrilateral lying in a plane not containing 0.)

Let us now drop the condition that the p_{i} are minimal among linearly dependent families, assuming only that none of them is a nonnegative multiple of another (i.e., that they determine distinct rays), and let S denote the union of the rays through 0 that they determine. Then one can show that a union of intervals ⋃_{i} [0, λ_{i} p_{i}] is relatively convex in S if and only if (25) holds for each linearly dependent family of ≥ 3 of the p_{i} which is minimal among linearly dependent families, and which has the property that the unique linear relation it satisfies has exactly one coefficient c_{j} of different sign from the rest. (The reduction to the minimal-linearly-independent-family case can be gotten by a recursive application of Carathéodory’s Theorem, with p_{0} = 0, within subspaces spanned by successively smaller subsets of {p_{1}, …, p_{n}}.)

As an example of the sort of lattice one gets, let us drop the central-symmetry and length ≤ 1 conditions from our snowflake construction, writing −p_{1}, −p_{2}, −p_{3} as p_{4}, p_{5}, p_{6}, and considering general sets

λ_{1} c.h.(0, p_{1}) ∪ … ∪ λ_{6} c.h.(0, p_{6}) (λ_{1}, …, λ_{6} ∈ [0, ∞]).

Then setting a_{i} = λ_{i}^{−1}, we get six conditions for relative convexity, namely

a_{2} ≤ a_{1} + a_{3}, a_{3} ≤ a_{2} + a_{4}, …, a_{1} ≤ a_{6} + a_{2},

each corresponding to the fact that one of our six rays lies in the cone spanned by its two immediate neighbors.

Our snowflake example showed that a lattice RelConv(S)_{≥{0}} could contain a 3-generator sublattice with an infinite descending chain. Let me sketch an example with an infinite ascending chain. In ℝ^{2}, let

p_{1} = (0, 3), p_{2} = (1, 2), p_{3} = (2, 1), p_{4} = (3, 0).

(Or for a more abstract description, take any four points of ℝ^{n} in arithmetic progression, on a line not passing through the origin.) Define

S_{i} = c.h.(0, p_{i}) and S = S_{1} ∪ S_{2} ∪ S_{3} ∪ S_{4}.

Now let L be the sublattice of RelConv(S)_{≥{0}} generated by

x_{1} = S_{1} ∪ (S_{2}/2), y = S_{2} ∪ S_{3}, x_{2} = (S_{3}/2) ∪ S_{4}.

It is easy to see from a sketch that, starting with x_{1} ∧ y, if we alternately apply (− ∨ x_{2}) ∧ y and (− ∨ x_{1}) ∧ y, we obtain an infinite ascending chain of sets.

Is the existence of infinite chains in sublattices of RelConv(S) generated by few elements limited to cases where S ≠ ℝ^{n}, or does it also occur in lattices Conv(ℝ^{n})? To get a large part of the answer without any computation, recall that RelConv(S) is isomorphic to the lattice of convex hulls in ℝ^{n} of subsets of S. This lattice, which we shall here denote L_{S}, is a subset but not a sublattice of Conv(ℝ^{n}); however, in cases like those considered above, where S is a finite union of convex sets, S = S_{1} ∪ ⋯ ∪ S_{m}, we can write the operations of this lattice as “polynomial operations” in those of Conv(ℝ^{n}). Namely, temporarily writing ∧_{S} and ∨_{S} for the operations of L_{S}, and ∧ and ∨ for those of Conv(ℝ^{n}), we get, for x, y ∈ L_{S},

x ∨_{S} y = x ∨ y,  x ∧_{S} y = ⋁_{i} (x ∧ y ∧ S_{i}).  (26)

Hence if the sublattice of L_{S} generated by elements y_{1}, …, y_{k} has an infinite ascending or descending chain (or any other specified join-sublattice), so will the sublattice of Conv(ℝ^{n}) generated by y_{1}, …, y_{k}, S_{1}, …, S_{m}. In the case of our “snowflake lattice”, the y ’s and the S ’s happen to be the same, so we immediately conclude that the sublattice of Conv(ℝ^{2})_{≥{0}} generated by those three elements has an infinite descending chain. From our example with an ascending chain, the best conclusion this general argument gives is that the sublattice of Conv(ℝ^{2})_{≥{0}} generated by the six elements S_{1}, S_{2}, S_{3}, S_{4}, S_{2}/2 and S_{3}/2 has an infinite ascending chain. Nevertheless, a little diagram-drawing shows that in this case, the sublattice of Conv(ℝ^{2})_{≥{0}} generated by the three elements c.h.(x_{1}), c.h.(x_{2}) and c.h.(y) shows essentially the same behavior as our lattice of relatively convex subsets.

I do not know any examples of 3-generator lattices of convex sets (relative or absolute) that have both infinite ascending and descending chains, or that have infinite antichains. On the other hand, it is not hard to show that the 4-generator sublattice of Conv(ℝ^{2}) generated by the diameters of a regular octagon has all three.

13. Notes on related work on relatively convex sets, and some further observations

The lattices RelConv(S) are examples of what are known as convex geometries; for the definition, and for results on these, see [7], [1].

Huhn [11] looked briefly at lattices of relatively convex sets determined by finite sets S ⊆ ℝ^{n} for the purpose of “approximating” Conv(ℝ^{n}) by finite lattices, and for the same purpose he considered in [10, §2] the dual construction, namely the lattice of those subsets of ℝ^{n} which can be represented as intersections of members of a given finite set of closed half-spaces. Not surprisingly in view of the dual natures of these two sorts of relativization, he found that lattices of the latter sort satisfied the identities of the form D_{n}^{op} that he had obtained in the nonrelativized lattice, but not those of the form D_{n} (compare Proposition 25 and Lemma 26 above).

In [19] it is shown that for every finite lattice L which is “lower bounded” (a strengthening of join semidistributive), there exist an n and a finite subset S ⊆ ℚ^{n} such that L is embeddable in the sublattice of Conv(ℝ^{n}) generated by {{p} | p ∈ S}. Clearly, such an embedding will send all members of L to convex polytopes; if we let S′ denote the union of the vertex-sets of this finite set of polytopes, it is not hard to see that we get an embedding of L in RelConv(S′). Whether embeddability in the lattice of relatively convex sets of a finite subset of ℝ^{n} holds not only for lower bounded lattices, but for all join-semidistributive lattices, is an open question [19, Problem 1], cf. [1, Problem 3]. It is also shown in [19] that every lattice can be embedded in the lattice of convex subsets of some infinite-dimensional vector space (over an arbitrary totally ordered division ring). Hence, in particular, lattices of the latter sort need not satisfy any nontrivial lattice identities. In contrast, the lattice of subspaces of any vector space satisfies the modular identity and others.

Here is another embedding result using not necessarily finite-dimensional vector spaces, although it may be seen that the connection with convexity is somewhat artificial, based on the fact that subspaces are in particular convex sets; it is essentially a result on lattices of “relative subspaces”. (Since most of the results in this note concerned subspaces of ℝ^{n}, we gave our basic definitions for that case only, but we shall use them here without that restriction.)

Lemma 28. Let V be a real vector space and B a basis of V. For each pair of distinct elements a, b ∈ B, let S_{a,b} be the subspace of V spanned by a − b. Define the set

S = ⋃_{a,b} S_{a,b} ⊆ V.

Then the following lattices are isomorphic:

(a) The lattice Equiv(B) of all equivalence relations on B (ordered by inclusion).

(b) The lattice RelSubsp(S) consisting of all sets of the form S ∩ U where U is a subspace of V. (Here meet is given by intersection; join by taking the subspace spanned by the union of the sets in question and intersecting it with S.)

(b′) The lattice of all subspaces of V spanned by subsets of S. (The join of two such subspaces W_{1}, W_{2} is W_{1} + W_{2}, the meet is the span of W_{1} ∩ W_{2} ∩ S.)

(c) The sublattice of RelConv(S) consisting of all elements thereof that are unions of subspaces S_{a,b}.

(c′) The lattice of all subsets of V which are convex hulls of unions of subspaces S_{a,b}. (Join as in Conv(V), while the meet of x_{1}, x_{2} is c.h.(x_{1} ∩ x_{2} ∩ S)).

Proof. That the sets described in (b), (b′), (c) and (c′), ordered by inclusion, form lattices, with meet and join as described, is immediate.

Note that the convex hull of a union of subspaces of V is the sum of those subspaces, and that the intersection of S with a subspace is always a union of certain of the S_{a,b}. From this it is easily seen that (b) and (c) are not merely, as asserted, isomorphic, but equal, and likewise (b′) and (c′). It is also clear that (b) is isomorphic to (b′), via the “span of” map in one direction and the operator − ∩ S in the other. So these four lattices are isomorphic; to complete the proof we shall describe an isomorphism between the lattice Equiv(B) of (a) and the lattice of (b′).

Given an equivalence relation R ∈ Equiv(B), let φ(R) be the subspace of V spanned by all elements a − b with (a, b) ∈ R, which by definition belongs to (b′), while given a subspace W ⊆ V spanned by a subset of S, let ψ(W) = {(a, b) | a, b ∈ B, a − b ∈ W }, which it is easy to check is an equivalence relation on B. The maps φ and ψ are clearly isotone, and from the definition of (b′), we see that φψ is the identity function thereof; moreover, for any R ∈ Equiv(B) it is clear that ψ(φ(R)) ≥ R, so it remains to prove the reverse inequality.

So suppose that (a, b) ∈ ψ(φ(R)), i.e., that a − b ∈ φ(R). From the definition of φ(R) it is easy to see that for each R-equivalence class C ⊆ B, the sum of the coefficients of all members of C in any element of φ(R) is zero. But the only way this can hold for a − b is if a and b are in the same equivalence class, i.e., (a, b) ∈ R, as required to complete our proof. □

Pudlák and Tůma [18] have shown that any finite lattice L embeds in Equiv(X) for some finite set X. Hence by the above lemma, for every such L one can find an n and a subset S ⊆ ℝ^{n} such that L is embeddable in RelConv(S). It would be interesting to know whether this same conclusion can be proved without using the deep result of [18]. So far as is known, the embedding of L in a lattice Equiv(X) may require a set X whose cardinality is enormous compared with that of L (though [14] and [12] somewhat improve the upper bound of [18]); but it is conceivable that one can do better with embeddings in lattices of relatively convex sets.

Our final remark will concern the observations at the end of the preceding section on the form of the lattice operations of RelConv(S) when S is a finite union of convex sets S_{1} ∪ ⋯ ∪ S_{m}. Let us put these in a more general context. (Readers allergic to category theory may ignore this discussion.)

Let Lattice denote the category of all lattices, objects of which we will here write L = (|L|, ∨, ∧), distinguishing between the lattice L and its underlying set |L|. For m a natural number, let Lattice^{m-pt} denote the category of lattices with m distinguished elements, i.e., of systems (|L|, ∨, ∧, S_{1}, …, S_{m}) such that (|L|, ∨, ∧) is a lattice and S_{1}, …, S_{m} ∈ |L|, and where a morphism between such systems means a lattice homomorphism which respects the ordered m-tuple of distinguished elements. Let us, finally, write Lattice^{int} for the category of objects (|L|, ∨, ∧, int), where (|L|, ∨, ∧) is a lattice, and int is an “interior operator” (the dual of a closure operator), that is, a map |L| → |L| satisfying

int(x) ≤ x,  x ≤ y ⟹ int(x) ≤ int(y),  int(int(x)) = int(x).

(Here by “ u ≤ v” we of course mean u = u ∧ v, equivalently, u ∨ v = v.) The morphisms of Lattice^{int} will be the lattice homomorphisms respecting this additional operation.

We can define a functor Lattice^{m-pt} → Lattice^{int} taking each object (|L|, ∨, ∧, S_{1}, …, S_{m}) to the object (|L|, ∨, ∧, int_{(S_{i})}), with interior operator defined by

int_{(S_{i})}(x) = ⋁_{i}(x ∧ S_{i}),

and another functor Lattice^{int} → Lattice, taking each object (|L|, ∨, ∧, int) to the object (|L|^{int}, ∨, ∧^{int}), where

|L|^{int} = {x ∈ |L| | x = int(x)} and x ∧^{int} y = int(x ∧ y).

(It is not hard to verify that the join operation of L carries |L|^{int} into itself.)

We now see that if we take L = Conv(ℝ^{n}), and let S_{1}, …, S_{m} be any m elements of this lattice, then the composite of the above two functors, applied to (|L|, ∨, ∧, S_{1}, …, S_{m}), gives precisely the lattice we named L_{S} ≅ RelConv(S), for S = S_{1} ∪ ⋯ ∪ S_{m}.

The constructions given by the above functor Lattice^{int} → Lattice, and its dual, with a closure operator replacing the interior operator, are well-known, if not in this functorial form. I do not know whether the construction Lattice^{m-pt} → Lattice^{int} (and its variant with lattices replaced by complete lattices and the specified finite family by an arbitrary family) has been considered.

Stepping back a little further, we may observe that the lattice Conv(ℝ^{n}) arises as the fixed set of the closure operator c.h.(−) on the lattice 2^{ℝ^{n}} of subsets of ℝ^{n}, so that the construction of the mutually isomorphic lattices L_{S} and RelConv(S) can be seen as arising from the interaction of the closure operator c.h.(−) and the interior operator − ∩ S on 2^{ℝ^{n}}. Again, this situation can be made into a general construction.

The reader familiar with the concept of representable algebra-valued functors ([3, Chapter 9] or [4, §1, §8]) will be happy to observe that all the functors of the above discussion are representable.

14. A question

The referee has pointed out that some properties of the lattice of convex sets are known to change if the base field ℝ is replaced by another ordered field (e.g., ℚ), but that the arguments of §§1-6 look as though they should work over any ordered field; perhaps even any ordered division ring. I have the same feeling, but as an amateur in the area, I will leave this question to others. (I do not know whether the theorems of Helly, Carathéodory and Radon hold in that context, nor how much of what I have justified as geometrically evident may rely on properties of the reals.)

A straightforward generalization of these results could not, of course, extend to §7, on compact convex sets, since over an ordered field which is not locally compact, the only nonempty compact convex sets are the singletons. One might be able to prove results like those of that section with “compact” replaced by “closed and bounded”, but new proofs would be needed, since the theorem of Minkowski we used there is not true in that context. Later sections depend to varying degrees on that one.

References

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University of California, Berkeley, CA 94720-3840, USA

E-mail address: gbergman@math.berkeley.edu