On the size of Heyting Semi-Lattices and Equationally Linear Heyting Algebras ∗
Peter Freyd [email protected]
July 17, 2017
The theory of Heyting Semi-Lattices, Finitely generated hsls are finite. hsls for short, is obtained by adding to the theory of meet-semi-lattices-with-top a The proof takes a while. First, say that binary operation whose values are denoted an hsl is local if there’s an element m < 1 x → y characterized by such that for all x < 1 it is the case that x ≤ m. (The local hsls are the “subdi- x ≤ y → z iff x ∧ y ≤ z rectly irreducibles” of the subject for those who know what that ugly phrase means.) The The easiest equational treatment uses the eponymous example is obtained by starting symmetric version, the double arrow rather with the hsl of open sets of a space, choos- than the single. The binary operation x ↔ y ing a point, identifying two open sets iff they is characterized by agree when restricted to some neighborhood x ≤ y ↔ z iff x ∧ y = x ∧ z of the point. We’ll call a local quotient a localization. Each may be defined in terms of the other: Every hsl is embedded into the product of x → y = x ↔ (x ∧ y) its localizations. x ↔ y = (x → y) ∧ (y → x) For two elements a and b we need to find a The ↔ operation may be equationally localization in which a and b remain distinct. characterized by Define m = a ↔ b. It suffices to find a local- ization in which m remains different from 1. x ↔ 1 = x = 1 ↔ x Reduce by a maximal congruence that does x ↔ x = 1 just that. In the resulting quotient hsl m has the property that for any non-trivial con- x ∧ (y ↔ z) = x ∧ ((x ∧ y) ↔ (x ∧ z)) [1] gruence it is the case that m ≡ 1. Let k be Any finite distributive lattice has a unique an element strictly less than 1. Define a con- Heyting structure. Every finite Heyting semi- gruence by x ≡ y iff x ∧ k = y ∧ k. (The lattice has, of course, a unique lattice struc- third defining equation for ↔ makes it easy ture and it is necessarily distributive. to see that this a congruence.) Since ≡ is a ∗ This ms was born in 2002. The first appendix was added non-trivial congruence (k ≡ 1) it must be the in 2005, the first footnote in 2013, the subscorings in 2015 and case that m ≡ 1, that is m ∧ k = k. In the addendum in 2017. 1[] See 2nd appendix for subscorings. First x and y meet other words, k ≤ m. x ↔ y in the same way: x∧(x ↔ y) = x∧((x∧x) ↔ (x∧y)) = x∧(x ↔ (x∧y)) = x∧((x∧1) ↔ (x∧y)) = x∧(1 ↔ y) = x∧y and similarly y ∧ (x ↔ y) = x ∧ y. Hence for any z ≤ x ↔ y Assume now that we know that all hsls we have z ∧ x = z ∧ (x ↔ y) ∧ y = z ∧ x ∧ y and similarly generated by n elements are finite. Let An be z ∧ y = z ∧ (x ↔ y) ∧ y = z ∧ x ∧ y. Finally, if z ∧ x = z ∧ y then z ∧ (x ↔ y) = z ∧ ((z ∧ x) ↔ (z ∧ y)) = z ∧ 1 = z, that the number of elements in the free hsl on n is, z ≤ x ↔ y. generators. HEYTING SEMI LAATTICES
A local hsl on n + 1 generators has at It’s easy to see that the free hsl on one most 1+ An elements. generator has 2 elements. We’ll discover be- low that the free on two generators has If m is removed the complement is a sub- hsl 18 elements and is isomorphic the the prod- algebra, because in a local hsl x → y = m uct 2×3×3 where the numbers name totally only if y = m (and x = 1) and x∧y = m only ordered sets. if either x or y is m. One of the n+1 genera- tors must be m. The complement is therefore Define an equationally linear Heyting generated by n of its elements. semi-lattice, or elhsl, to be an hsl that satisfies the further equation: Hence for an hsl on n+1 generators there are only a finite number of isomorphism types ((x → y) → z) ∧ ((y → x) → z)) = z of localizations. From the finiteness of the generating set we know that there are only Any totally ordered set with a top element finitely many maps into any finite hsl. We is a elhsl. And any local elhsl is linear, may conclude that there are only finitely that is, totally ordered (specialize to the case many localizations. Which is quite enough to that z = m). Hence an hsl is equationally force the hsl on n+1 generators to be finite. linear iff it can be embedded in a product of (If the join operation and a bottom are linear hsls. added to the structure, that is, if we con- For a finite set, G, of variables let F (G) sider not hsls but full-fledged Heyting alge- denote the free elhsl on G. F (∅) = {1}. bras, then the free ha on one generator is in- Theorem: finite. For a quick proof, partially order the ∼ Y positive integers by taking x <
2 HEYTING SEMI LAATTICES map from H to H/k that sends x to x ∧ k as well call them linear has. (The defini- is. That is, H/k may be viewed as a quotient tion of linear has can be made a bit simpler: structure of H. (Indeed, if H is finite or, more they are those has that satisfy the equation generally, satisfies the descending chain condi- (x → y) ∨ (y → x) = 1.) All of which al- tion, then every quotient structure so arises.) lows us to prove that the free linear Heyting The product decompositions of H are of the algebra on G is constructible as: form H/k1 × H/k2 × · · · where the sequence Y F (S)⊥ {k1, k2,...} is a “partition of unity”, that is, a sequence of elements in H above the bottom S⊆G element whose join is 1 and whose pairwise and that the order of the free linear ha on n meets are each the bottom element. In the fi- generators is: nite case there is a unique largest such. That is, any finite hsl uniquely factors as a prod- (f(n) + 1)f(n) uct of “prime” algebras. (All of this is easy in The first few values are 2, 6, 342, the distributive lattice context. It continues 137186159382. (The same asymptotic expres- to work in the ha context. And even in the sion works as for f(n) with C approximately hsl context.) 4.1545.) Given a finite set, G, let 0 denote the el- V (There is something of a converse of the ement a∈G a in F (G) (0 is the bottom el- fact that join is definable in elhsls: if the join ement). For any element x let ¬x denote operation is definable in a variety of hsls then x → 0. For each subset S ⊂ G define necessarily it is a variety of equationally linear ^ ^ hsls. See Appendix. There aren’t many such tS = ( ¬a) ∧ ( ¬¬a) varieties, by the way. Each is determined by a∈G\S a∈S the sizes of its local members, that is, each The family {tS}S⊂G, S6=G is a partition of is determined by the smallest n, if any, for unity. To see that, it suffices to show in any Wn which the equation 1 = i=1(xi → xi+1) is local elhsl generated by G, that there is a satisfied.) unique S ⊂ G such that t names the top el- S The number of isomorphism types of ement and for all other proper subsets of G, elhsls of order n already appears in Sloane t names the bottom element. S as A050318, albeit shifted by 1, credited to When we regard F (G)/tS as a finitely pre- Christian G. Bower whose description of the sented elhsl (where tS = 1 is the unique re- sequence there is: lation) it is easily seen to be isomorphic to F (S)⊥. Number of ways to write n as an It is easy to verify that any local hsl on mterm, where an mterm is an un- two generators is totally ordered, hence any ordered sum which is either 2, or 1 hsl on two generators is linear. Necessarily + an unordered product of mterms. the free hsl on two generators is the same as Formula: Shifts left under trans- the free elhsl on two generators. form T where T a has Dirichlet g.f.: ∞ In any local elhsl Y (1/(1 − 1/ns)a(n)) ((x → y) → y) ∧ ((y → x) → x) n=1 can be verified to satisfy the equations of a I would rewrite that as join operation (because, of course, it can be verified to be the join operation), hence con- Number of ways to write n as an tinues to satisfy those equations in any elhsl. mterm, where an mterm is the sum If we add a bottom constant then a elhsl is 1 + an unordered product (possibly a full-fledged Heyting algebra and we could empty) of mterms.
3 HEYTING SEMI LAATTICES
The first few entries are: 1, 1, 1, 2, 2, 3, algebras (any finite hsl is automatically an 3, 5, 6, 8, 8, 12, 12, 15, 17, 23, 23, 31, 31, ha). But unlike the case with A006982, there 41, 44, 52, 52, 69, 72, 84, 90, 108, 108, 135, is no happy translation into distributive lat- 135, 161, 169, 192, 198, 246, 246, 277, 289, tices. Perhaps the least unhappy: A050318 342, 342, 404, 404, 464, 491, 543, 543, 644, shifted by 1 is the number of isomorphism 650, 734, 757, 853, 853, 978, 994, 1123, 1154, types of distributive lattices in which any el- 1262, 1262, ... ement below a coprime is itself a coprime. Which, of course, is the same as the number I’ll use the convention that 2 names the of distributive lattices in which any element mterm 1+1 (where the second 1 is the empty above a prime is itself a prime. product), 3 names 1+2, 4 names 1+3, etc. The first number that may be written as an Finally, Bower’s sequence A050365, can mterm in more that one way is five, to wit, be interpreted to count the number of 5 = 1 + 4 and 1 + (2×2). There remain only anti-symmetric elhsls (and elhas), to two ways for six, to wit, 6 = 1 + 5 and 1 + wit, those for which all automorphisms are (1 + (2×2)). For seven there are three ways: one. His description (emphasis added): 7 = 1 + 6, 1 + (1 + (1 + (2×2))) and 1 + (2×3). For thirteen there are twelve ways (using a further simplification of notation): 13 = 1 + Ways to write n as an identity 12, 9+(2×2), 7+(2×3), 6+(2×4), 5+(2×2×2), mterm, where an identity mterm is 4 + (3×3), 3 + (2×5), 3 + (2×(1 + (2×2)))), an unordered sum which is either 1 + (2×(2 + (2×2))), 1 + (2×2×3), 1 + (3×4), 2, or 1 + an unordered product of 1 + (2×6). distinct identity mterms. Formula: Shifts left under transform T where The connection with s is this: Ev- elhsl T a has Dirichlet g.f.: ery prime elhsl is the lifting of an elhsl of one fewer element, which elhsl is, as already ∞ Y pointed out, uniquely an unordered product (1 + 1/ns)a(n) of prime elhsls. The argument is as follows. n=1 If 0 is a prime element of the given elhsl, that is, if 0 is not the meet of two larger ele- ments then—since we are in a finite lattice— The first few terms are 1, 1, 1, 1, 1, 2, 2, there is an element just on top of 0 that is 3, 3, 4, 4, 6, 6, 8, 9, 11, 11, 15, 15, 19, 21, below all other elements above 0, hence the 25, 25, 33, 33, 39, 42, 50, 50, 63, 63, 74, 78, elhsl is a lifting of an elhsl of one fewer el- 89, 91, 110, 110, 125, 131, 152, 152, 181, 181, ement. If 0 is not prime we will argue that the 206, 217, 242, 242, 285, 286, 322, 333, 372, elhsl is not prime. It suffices to find a non- 372, 428, 432, 486, 501, 551, 551, 636, 636, trivial partition of unity. Let a > 0 and b > 0 699, 724, 799 ... be such that a ∧ b = 0. Then ¬a ∧ ¬¬a = 1 There is yet another interpretation. Re- (it suffices to verify in any totally ordered hsl that (x → y) ∧ ((x → y) → y) = 1). ¬a > 0 call that in formal set theory the rank of a since ¬a ≥ b. ¬¬a > 0 since ¬¬a ≥ a. Hence set is defined, recursively, as the first ordinal larger than the rank of any of its elements. A050318 counts the number of prime elhsls. For a set of finite rank we may define its But the number of prime elhsls of order n Bower-rank as the first ordinal larger than is equal to the number of all elhsls of order the product of the Bower-ranks of its elements. n−1 since each is the lifting of a unique elhsl of one fewer element. Then A050365 counts the number of sets of given Bower-rank. If we extend these defi- As with A006982, we may describe nitions to “mull-sets” then A050318, besides A050318 as counting the number of isomor- counting mterms, elhsls and elhas, counts phism classes of equationally linear Heyting the number of multi-sets of given Bower-rank.
4 HEYTING SEMI LAATTICES
Appendix 1 B-element is a C-element. Clearly C ∪ {1} and B are each closed un- The following are equivalent conditions on der the arrow operation. Clearly if c ∈ C and a variety of hsls: b ∈ B then c → b = 1 and b → c = c. Hence a: All members are elhsls; C∪B is a subalgebra. It lies in the variety un- der discussion. B is the set of upper bounds b: All local members are totally ordered; of the finite set C. It does not have a least c: All members with a bottom are Heyting element. algebras. We have already proved that a and b are equivalent and that they imply c. We need Appendix 2 only show that c implies b. We’ll attack the contrapositive: if a variety of hsls has a local x ∧ (x ↔ y) y ∧ (x ↔ y) member that is not totally ordered then it has a member with a finite subset of elements x ∧ ((x ∧ x) ↔ (x ∧ y)) y ∧ ((y ∧ x) ↔ (y ∧ y)) whose set of upper bounds does not have a x ∧ (x ↔ (x ∧ y)) y ∧ ((y ∧ x) ↔ y) bottom. x ∧ ((x ∧ 1) ↔ (x ∧ y)) y ∧ ((y ∧ x) ↔ (y ∧ 1)) If there’s a local member that is not to- tally ordered we may cut down to the finite x ∧ (1 ↔ y) y ∧ (x ↔ 1) subalgebra generated by m and a pair of in- x ∧ y == y ∧ x comparable elements. We may then chose a smallest such example. We’ll call it A. Let m0 be the join of all the elements less z ≤ x ↔ y than m. If m0 < m then A/m [2] is still local z ∧ x z ∧ y and still not totally ordered. Hence m is the join of the smaller elements. z ∧ (x ↔ y) ∧ y z ∧ (x ↔ y) ∧ x Let P be the countable cartesian power of z ∧ x ∧ y == z ∧ y ∧ x A. We will treat its elements as sequences of A-elements. We carve out a particular subal- gebra as the (disjoint) union of two subsets C z ∧ x = z ∧ y and B. C is the set of all constant sequences with constant value less than m. B is the set z ∧ (x ↔ y) of all sequences that are equal to 1 except for z ∧ ((z ∧ x) ↔ (z ∧ y)) a finite number of ms. Note that every ele- ment of C is less than every element of B. z ∧ ((z ∧ x) ↔ (z ∧ x))
Clearly both C and B are closed under the z ∧ 1 meet operation, and the last remark makes it clear that the meet of a C-element and a z
2[] At the end of page 2 it’s defined as the set {x ∈ A | x ≤ m} with the induced hsl structure.
5 HEYTING SEMI LAATTICES
Addendum equation. So let y be less than m, take x = z = m to obtain (x → z) ∧ ((x → y) → z) = The theory of hsls is an example of an (m → m) ∧ ((m → y) → m) = 1 ∧ (y → m) equational theory with a unique maximal = 1 ∧ 1 6= z. consistent extension,[3] and only one equation is needed to achieve that extension: Finally, any hsl with a bottom that satis- fies this equation is a Boolean algebra.[5] (x → z) ∧ ((x → y) → z) = z Following the lead of footnote [17] in www.math.upenn.edu/~pjf/amplifications.pdf Its consistency is established by checking wherein abelian group theory without zero is its correctness on a two-element hsl.[4] Since named—in imitation of Jacobson’s rngs—the any non-trivial hsl contains a two-element theory of grups we can call such algebras subalgebra (to wit, {x, 1} for any x < 1) any bolean. equation is consistent with the theory of hsls Better known is its upside-down version. iff it holds for a two-element hsl. Consider the join-semilattice of finite sub- We establish that this one new equation sets (of an infinite set). Add the relative- suffices for all consistent equations by es- complement operation, A \ B, and the dual tablishing that any hsl that satisfies it can of (x → z) ∧ ((x → y) → z) = z to obtain be embedded in a cartesian power of the (C \ A) ∪ (C \ (B \ A)) = C. If this doesn’t two-element hsl. We know that any hsl is look familiar, use AB to denote A \ (A \ B) embedded in the product of its localizations, and A+B to denote (A\B)∪(B \A). It’s not hence we finish by showing that any local hsl called a boolan algebra. It’s called a Boolean with more than two elements fails the new rng. J J
Available at http://www.math.upenn.edu/~pjf/Heyting.pdf
3[] See footnote [35] in http://www.math.upenn.edu/~pjf/analysis.pdf 4[] The equation easily holds when z = 1 and (x → 0) ∧ ((x → y) → 0) = 0 easily holds when either x = 1 5[] It’s clearly an elhsl hence has a join operation. (Indeed, or y = 1. And that leaves only (0 → 0) ∧ ((0 → 0) → 0) = x ∨ y is even more easily defined just as (x → y) → y.) Add 1 ∧ (1 → 0) = 0. a bottom 0 and use x → 0 for the complement.
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